Suffix tree: Difference between revisions
SqrtNegInf (talk | contribs) m →{{header|Perl}}: Fix link: Perl 6 --> Raku |
SqrtNegInf (talk | contribs) m →{{header|Sidef}}: Fix link: Perl 6 --> Raku |
||
Line 1,065:
=={{header|Sidef}}==
{{trans|
<lang ruby>func suffix_tree(Str t) {
suffix_tree(^t.len -> map { t.substr(_) })
|
Revision as of 14:53, 15 March 2020
A suffix tree is a data structure commonly used in string algorithms.
Given a string S of length n, its suffix tree is a tree T such that:
- T has exactly n leaves numbered from 1 to n.
- Except for the root, every internal node has at least two children.
- Each edge of T is labelled with a non-empty substring of S.
- No two edges starting out of a node can have string labels beginning with the same character.
- The string obtained by concatenating all the string labels found on the path from the root to leaf i spells out suffix S[i..n], for i from 1 to n.
Such a tree does not exist for all strings. To ensure existence, a character that is not found in S must be appended at its end. The character '$' is traditionally used for this purpose.
For this task, build and display the suffix tree of the string "banana$". Displaying the tree can be done using the code from the visualize a tree task, but any other convenient method is accepted.
There are several ways to implement the tree data structure, for instance how edges should be labelled. Latitude is given in this matter, but notice that a simple way to do it is to label each node with the label of the edge leading to it.
The computation time for an efficient algorithm should be , but such an algorithm might be difficult to implement. An easier, algorithm is accepted.
C#
<lang csharp>using System; using System.Collections.Generic;
namespace SuffixTree {
class Node { public string sub; // a substring of the input string public List<int> ch = new List<int>(); // vector of child nodes
public Node() { sub = ""; }
public Node(string sub, params int[] children) { this.sub = sub; ch.AddRange(children); } }
class SuffixTree { readonly List<Node> nodes = new List<Node>();
public SuffixTree(string str) { nodes.Add(new Node()); for (int i = 0; i < str.Length; i++) { AddSuffix(str.Substring(i)); } }
public void Visualize() { if (nodes.Count == 0) { Console.WriteLine("<empty>"); return; }
void f(int n, string pre) { var children = nodes[n].ch; if (children.Count == 0) { Console.WriteLine("- {0}", nodes[n].sub); return; } Console.WriteLine("+ {0}", nodes[n].sub);
var it = children.GetEnumerator(); if (it.MoveNext()) { do { var cit = it; if (!cit.MoveNext()) break;
Console.Write("{0}+-", pre); f(it.Current, pre + "| "); } while (it.MoveNext()); }
Console.Write("{0}+-", pre); f(children[children.Count-1], pre+" "); }
f(0, ""); }
private void AddSuffix(string suf) { int n = 0; int i = 0; while (i < suf.Length) { char b = suf[i]; int x2 = 0; int n2; while (true) { var children = nodes[n].ch; if (x2 == children.Count) { // no matching child, remainder of suf becomes new node n2 = nodes.Count; nodes.Add(new Node(suf.Substring(i))); nodes[n].ch.Add(n2); return; } n2 = children[x2]; if (nodes[n2].sub[0] == b) { break; } x2++; } // find prefix of remaining suffix in common with child var sub2 = nodes[n2].sub; int j = 0; while (j < sub2.Length) { if (suf[i + j] != sub2[j]) { // split n2 var n3 = n2; // new node for the part in common n2 = nodes.Count; nodes.Add(new Node(sub2.Substring(0, j), n3)); nodes[n3].sub = sub2.Substring(j); // old node loses the part in common nodes[n].ch[x2] = n2; break; // continue down the tree } j++; } i += j; // advance past part in common n = n2; // continue down the tree } } }
class Program { static void Main() { new SuffixTree("banana$").Visualize(); } }
}</lang>
- Output:
+ +-- banana$ +-+ a | +-+ na | | +-- na$ | | +-- $ | +-- $ +-+ na | +-- na$ | +-- $ +-- $
C++
<lang cpp>#include <functional>
- include <iostream>
- include <vector>
struct Node {
std::string sub = ""; // a substring of the input string std::vector<int> ch; // vector of child nodes
Node() { // empty }
Node(const std::string& sub, std::initializer_list<int> children) : sub(sub) { ch.insert(ch.end(), children); }
};
struct SuffixTree {
std::vector<Node> nodes;
SuffixTree(const std::string& str) { nodes.push_back(Node{}); for (size_t i = 0; i < str.length(); i++) { addSuffix(str.substr(i)); } }
void visualize() { if (nodes.size() == 0) { std::cout << "<empty>\n"; return; }
std::function<void(int, const std::string&)> f; f = [&](int n, const std::string & pre) { auto children = nodes[n].ch; if (children.size() == 0) { std::cout << "- " << nodes[n].sub << '\n'; return; } std::cout << "+ " << nodes[n].sub << '\n';
auto it = std::begin(children); if (it != std::end(children)) do { if (std::next(it) == std::end(children)) break; std::cout << pre << "+-"; f(*it, pre + "| "); it = std::next(it); } while (true);
std::cout << pre << "+-"; f(children[children.size() - 1], pre + " "); };
f(0, ""); }
private:
void addSuffix(const std::string & suf) { int n = 0; size_t i = 0; while (i < suf.length()) { char b = suf[i]; int x2 = 0; int n2; while (true) { auto children = nodes[n].ch; if (x2 == children.size()) { // no matching child, remainder of suf becomes new node n2 = nodes.size(); nodes.push_back(Node(suf.substr(i), {})); nodes[n].ch.push_back(n2); return; } n2 = children[x2]; if (nodes[n2].sub[0] == b) { break; } x2++; } // find prefix of remaining suffix in common with child auto sub2 = nodes[n2].sub; size_t j = 0; while (j < sub2.size()) { if (suf[i + j] != sub2[j]) { // split n2 auto n3 = n2; // new node for the part in common n2 = nodes.size(); nodes.push_back(Node(sub2.substr(0, j), { n3 })); nodes[n3].sub = sub2.substr(j); // old node loses the part in common nodes[n].ch[x2] = n2; break; // continue down the tree } j++; } i += j; // advance past part in common n = n2; // continue down the tree } }
};
int main() {
SuffixTree("banana$").visualize();
}</lang>
- Output:
+ +-- banana$ +-+ a | +-+ na | | +-- na$ | | +-- $ | +-- $ +-+ na | +-- na$ | +-- $ +-- $
D
<lang D>import std.stdio;
struct Node {
string sub = ""; // a substring of the input string int[] ch; // array of child nodes
this(string sub, int[] children ...) { this.sub = sub; ch = children; }
}
struct SuffixTree {
Node[] nodes;
this(string str) { nodes ~= Node(); for (int i=0; i<str.length; ++i) { addSuffix(str[i..$]); } }
private void addSuffix(string suf) { int n = 0; int i = 0; while (i < suf.length) { char b = suf[i]; int x2 = 0; int n2; while (true) { auto children = nodes[n].ch; if (x2 == children.length) { // no matching child, remainder of suf becomes new node. n2 = nodes.length; nodes ~= Node(suf[i..$]); nodes[n].ch ~= n2; return; } n2 = children[x2]; if (nodes[n2].sub[0] == b) { break; } x2++; } // find prefix of remaining suffix in common with child auto sub2 = nodes[n2].sub; int j = 0; while (j < sub2.length) { if (suf[i + j] != sub2[j]) { // split n2 auto n3 = n2; // new node for the part in common n2 = nodes.length; nodes ~= Node(sub2[0..j], n3); nodes[n3].sub = sub2[j..$]; // old node loses the part in common nodes[n].ch[x2] = n2; break; // continue down the tree } j++; } i += j; // advance past part in common n = n2; // continue down the tree } }
void visualize() { if (nodes.length == 0) { writeln("<empty>"); return; }
void f(int n, string pre) { auto children = nodes[n].ch; if (children.length == 0) { writefln("╴ %s", nodes[n].sub); return; } writefln("┐ %s", nodes[n].sub); foreach (c; children[0..$-1]) { write(pre, "├─"); f(c, pre ~ "│ "); } write(pre, "└─"); f(children[$-1], pre ~ " "); }
f(0, ""); }
}
void main() {
SuffixTree("banana$").visualize();
}</lang>
- Output:
┐ ├─╴ banana$ ├─┐ a │ ├─┐ na │ │ ├─╴ na$ │ │ └─╴ $ │ └─╴ $ ├─┐ na │ ├─╴ na$ │ └─╴ $ └─╴ $
Go
Vis function from Visualize_a_tree#Unicode. <lang go>package main
import "fmt"
func main() {
vis(buildTree("banana$"))
}
type tree []node
type node struct {
sub string // a substring of the input string ch []int // list of child nodes
}
func buildTree(s string) tree {
t := tree{node{}} // root node for i := range s { t = t.addSuffix(s[i:]) } return t
}
func (t tree) addSuffix(suf string) tree {
n := 0 for i := 0; i < len(suf); { b := suf[i] ch := t[n].ch var x2, n2 int for ; ; x2++ { if x2 == len(ch) { // no matching child, remainder of suf becomes new node. n2 = len(t) t = append(t, node{sub: suf[i:]}) t[n].ch = append(t[n].ch, n2) return t } n2 = ch[x2] if t[n2].sub[0] == b { break } } // find prefix of remaining suffix in common with child sub2 := t[n2].sub j := 0 for ; j < len(sub2); j++ { if suf[i+j] != sub2[j] { // split n2 n3 := n2 // new node for the part in common n2 = len(t) t = append(t, node{sub2[:j], []int{n3}}) t[n3].sub = sub2[j:] // old node loses the part in common t[n].ch[x2] = n2 break // continue down the tree } } i += j // advance past part in common n = n2 // continue down the tree } return t
}
func vis(t tree) {
if len(t) == 0 { fmt.Println("<empty>") return } var f func(int, string) f = func(n int, pre string) { children := t[n].ch if len(children) == 0 { fmt.Println("╴", t[n].sub) return } fmt.Println("┐", t[n].sub) last := len(children) - 1 for _, ch := range children[:last] { fmt.Print(pre, "├─") f(ch, pre+"│ ") } fmt.Print(pre, "└─") f(children[last], pre+" ") } f(0, "")
}</lang>
- Output:
┐ ├─╴ banana$ ├─┐ a │ ├─┐ na │ │ ├─╴ na$ │ │ └─╴ $ │ └─╴ $ ├─┐ na │ ├─╴ na$ │ └─╴ $ └─╴ $
J
Implementation:
<lang J>classify=: {.@> </. ]
build_tree=:3 :0
tree=. ,:_;_; if. 0=#y do. tree return.end. if. 1=#y do. tree,(#;y);0;y return.end. for_box.classify y do. char=. {.>{.>box subtree=. }.build_tree }.each>box ndx=.I.0=1&{::"1 subtree n=.#tree if. 1=#ndx do. counts=. 1 + 0&{::"1 subtree parents=. (n-1) (+*]&*) 1&{::"1 subtree edges=. (ndx}~ <@(char,ndx&{::)) 2&{"1 subtree tree=. tree, counts;"0 1 parents;"0 edges else. tree=. tree,(__;0;,char),(1;n;0) + ::]&.>"1 subtree end. end.
)
suffix_tree=:3 :0
assert. -.({:e.}:)y tree=. B=:|:build_tree <\. y ((1+#y)-each {.tree),}.tree
)</lang>
Task example:
<lang J> suffix_tree 'banana$' ┌──┬───────┬─┬──┬───┬─┬─┬──┬───┬─┬─┐ │__│1 │_│_ │2 │4│6│_ │3 │5│7│ ├──┼───────┼─┼──┼───┼─┼─┼──┼───┼─┼─┤ │_ │0 │0│2 │3 │3│2│0 │7 │7│0│ ├──┼───────┼─┼──┼───┼─┼─┼──┼───┼─┼─┤ │ │banana$│a│na│na$│$│$│na│na$│$│$│ └──┴───────┴─┴──┴───┴─┴─┴──┴───┴─┴─┘</lang>
The first row is the leaf number (_ for internal nodes).
The second row is parent index (_ for root node).
The third row is the edge's substring (empty for root node).
Visualizing, using showtree and prefixing the substring leading to each leaf with the leaf number (in brackets):
<lang J>fmttree=: ;@(1&{) showtree~ {: (,~ }.`('[','] ',~":)@.(_>|))each {.
fmttree suffix_tree 'banana$' ┌─ [1] banana$ │ ┌─ [2] na$ │ ┌─ na ────┴─ [4] $
────┼─ a ─────────┴─ [6] $
│ ┌─ [3] na$ ├─ na ────────┴─ [5] $ └─ [7] $
</lang>
Java
<lang Java>import java.util.ArrayList; import java.util.List;
public class SuffixTreeProblem {
private static class Node { String sub = ""; // a substring of the input string List<Integer> ch = new ArrayList<>(); // list of child nodes }
private static class SuffixTree { private List<Node> nodes = new ArrayList<>();
public SuffixTree(String str) { nodes.add(new Node()); for (int i = 0; i < str.length(); ++i) { addSuffix(str.substring(i)); } }
private void addSuffix(String suf) { int n = 0; int i = 0; while (i < suf.length()) { char b = suf.charAt(i); List<Integer> children = nodes.get(n).ch; int x2 = 0; int n2; while (true) { if (x2 == children.size()) { // no matching child, remainder of suf becomes new node. n2 = nodes.size(); Node temp = new Node(); temp.sub = suf.substring(i); nodes.add(temp); children.add(n2); return; } n2 = children.get(x2); if (nodes.get(n2).sub.charAt(0) == b) break; x2++; } // find prefix of remaining suffix in common with child String sub2 = nodes.get(n2).sub; int j = 0; while (j < sub2.length()) { if (suf.charAt(i + j) != sub2.charAt(j)) { // split n2 int n3 = n2; // new node for the part in common n2 = nodes.size(); Node temp = new Node(); temp.sub = sub2.substring(0, j); temp.ch.add(n3); nodes.add(temp); nodes.get(n3).sub = sub2.substring(j); // old node loses the part in common nodes.get(n).ch.set(x2, n2); break; // continue down the tree } j++; } i += j; // advance past part in common n = n2; // continue down the tree } }
public void visualize() { if (nodes.isEmpty()) { System.out.println("<empty>"); return; } visualize_f(0, ""); }
private void visualize_f(int n, String pre) { List<Integer> children = nodes.get(n).ch; if (children.isEmpty()) { System.out.println("- " + nodes.get(n).sub); return; } System.out.println("┐ " + nodes.get(n).sub); for (int i = 0; i < children.size() - 1; i++) { Integer c = children.get(i); System.out.print(pre + "├─"); visualize_f(c, pre + "│ "); } System.out.print(pre + "└─"); visualize_f(children.get(children.size() - 1), pre + " "); } }
public static void main(String[] args) { new SuffixTree("banana$").visualize(); }
}</lang>
- Output:
┐ ├─- banana$ ├─┐ a │ ├─┐ na │ │ ├─- na$ │ │ └─- $ │ └─- $ ├─┐ na │ ├─- na$ │ └─- $ └─- $
Kotlin
<lang scala>// version 1.1.3
class Node {
var sub = "" // a substring of the input string var ch = mutableListOf<Int>() // list of child nodes
}
class SuffixTree(val str: String) {
val nodes = mutableListOf<Node>(Node())
init { for (i in 0 until str.length) addSuffix(str.substring(i)) }
private fun addSuffix(suf: String) { var n = 0 var i = 0 while (i < suf.length) { val b = suf[i] val children = nodes[n].ch var x2 = 0 var n2: Int while (true) { if (x2 == children.size) { // no matching child, remainder of suf becomes new node. n2 = nodes.size nodes.add(Node().apply { sub = suf.substring(i) } ) children.add(n2) return } n2 = children[x2] if (nodes[n2].sub[0] == b) break x2++ } // find prefix of remaining suffix in common with child val sub2 = nodes[n2].sub var j = 0 while (j < sub2.length) { if (suf[i + j] != sub2[j]) { // split n2 val n3 = n2 // new node for the part in common n2 = nodes.size nodes.add(Node().apply { sub = sub2.substring(0, j) ch.add(n3) }) nodes[n3].sub = sub2.substring(j) // old node loses the part in common nodes[n].ch[x2] = n2 break // continue down the tree } j++ } i += j // advance past part in common n = n2 // continue down the tree } }
fun visualize() { if (nodes.isEmpty()) { println("<empty>") return }
fun f(n: Int, pre: String) { val children = nodes[n].ch if (children.isEmpty()) { println("╴ ${nodes[n].sub}") return } println("┐ ${nodes[n].sub}") for (c in children.dropLast(1)) { print(pre + "├─") f(c, pre + "│ ") } print(pre + "└─") f(children.last(), pre + " ") }
f(0, "") }
}
fun main(args: Array<String>) {
SuffixTree("banana$").visualize()
}</lang>
- Output:
┐ ├─╴ banana$ ├─┐ a │ ├─┐ na │ │ ├─╴ na$ │ │ └─╴ $ │ └─╴ $ ├─┐ na │ ├─╴ na$ │ └─╴ $ └─╴ $
Perl
<lang Perl>use strict; use warnings; use Data::Dumper;
sub classify {
my $h = {}; for (@_) { push @{$h->{substr($_,0,1)}}, $_ } return $h;
} sub suffixes {
my $str = shift; map { substr $str, $_ } 0 .. length($str) - 1;
} sub suffix_tree {
return +{} if @_ == 0; return +{ $_[0] => +{} } if @_ == 1; my $h = {}; my $classif = classify @_; for my $key (keys %$classif) { my $subtree = suffix_tree( map { substr $_, 1 } @{$classif->{$key}} ); my @subkeys = keys %$subtree; if (@subkeys == 1) { my ($subkey) = @subkeys; $h->{"$key$subkey"} = $subtree->{$subkey}; } else { $h->{$key} = $subtree } } return $h;
} print +Dumper suffix_tree suffixes 'banana$';</lang>
- Output:
$VAR1 = { '$' => {}, 'a' => { '$' => {}, 'na' => { 'na$' => {}, '$' => {} } }, 'banana$' => {}, 'na' => { 'na$' => {}, '$' => {} } };
Phix
<lang Phix>-- tree nodes are simply {string substr, sequence children_idx} enum SUB=1, CHILDREN=2
function addSuffix(sequence t, string suffix)
int n = 1, i = 1 while i<=length(suffix) do integer ch = suffix[i], x2 = 1, n2 while (true) do sequence children = t[n][CHILDREN] if x2>length(children) then -- no matching child, remainder of suffix becomes new node. t = append(t,{suffix[i..$],{}}) t[n][CHILDREN] &= length(t) return t end if n2 = children[x2] if t[n2][SUB][1]==ch then exit end if x2 += 1 end while -- find prefix of remaining suffix in common with child string prefix = t[n2][SUB] int j = 0 while j<length(prefix) do if suffix[i+j]!=prefix[j+1] then -- split n2: new node for the part in common t = append(t,{prefix[1..j],{n2}}) -- old node loses the part in common t[n2][SUB] = prefix[j+1..$] -- and child idx moves to newly created node n2 = length(t) t[n][CHILDREN][x2] = n2 exit -- continue down the tree end if j += 1 end while i += j -- advance past part in common n = n2 -- continue down the tree end while return t
end function
function SuffixTree(string s)
sequence t = {{"",{}}} for i=1 to length(s) do t = addSuffix(t,s[i..$]) end for return t
end function
procedure visualize(sequence t, integer n=1, string pre="")
if length(t)=0 then printf(1,"<empty>\n"); return; end if sequence children = t[n][CHILDREN] if length(children)=0 then printf(1,"- %s\n", {t[n][SUB]}) return end if printf(1,"+ %s\n", {t[n][SUB]}) integer l = length(children) for i=1 to l do puts(1,pre&"+-") visualize(t,children[i],pre&iff(i=l?" ":"| ")) end for
end procedure
sequence t = SuffixTree("banana$") visualize(t)</lang>
- Output:
+ +-- banana$ +-+ a | +-+ na | | +-- na$ | | +-- $ | +-- $ +-+ na | +-- na$ | +-- $ +-- $
Python
<lang python>class Node:
def __init__(self, sub="", children=[]): self.sub = sub self.ch = children
class SuffixTree:
def __init__(self, str): self.nodes = [Node()] for i in range(len(str)): self.addSuffix(str[i:])
def addSuffix(self, suf): n = 0 i = 0 while i < len(suf): b = suf[i] x2 = 0 while True: children = self.nodes[n].ch if x2 == len(children): # no matching child, remainder of suf becomes new node n2 = len(self.nodes) self.nodes.append(Node(suf[i:], [])) self.nodes[n].ch.append(n2) return n2 = children[x2] if self.nodes[n2].sub[0] == b: break x2 = x2 + 1
# find prefix of remaining suffix in common with child sub2 = self.nodes[n2].sub j = 0 while j < len(sub2): if suf[i + j] != sub2[j]: # split n2 n3 = n2 # new node for the part in common n2 = len(self.nodes) self.nodes.append(Node(sub2[:j], [n3])) self.nodes[n3].sub = sub2[j:] # old node loses the part in common self.nodes[n].ch[x2] = n2 break # continue down the tree j = j + 1 i = i + j # advance past part in common n = n2 # continue down the tree
def visualize(self): if len(self.nodes) == 0: print "<empty>" return
def f(n, pre): children = self.nodes[n].ch if len(children) == 0: print "--", self.nodes[n].sub return print "+-", self.nodes[n].sub for c in children[:-1]: print pre, "+-", f(c, pre + " | ") print pre, "+-", f(children[-1], pre + " ")
f(0, "")
SuffixTree("banana$").visualize()</lang>
- Output:
+- +- -- banana$ +- +- a | +- +- na | | +- -- na$ | | +- -- $ | +- -- $ +- +- na | +- -- na$ | +- -- $ +- -- $
Racket
See Suffix trees with Ukkonen’s algorithm by Danny Yoo for more information on how to use suffix trees in Racket.
<lang racket>#lang racket (require (planet dyoo/suffixtree)) (define tree (make-tree)) (tree-add! tree (string->label "banana$"))
(define (show-node nd dpth)
(define children (node-children nd)) (printf "~a~a ~a~%" (match dpth [(regexp #px"(.*) $" (list _ d)) (string-append d "`")] [else else]) (if (null? children) "--" "-+") (label->string (node-up-label nd))) (let l ((children children)) (match children ((list) (void)) ((list c) (show-node c (string-append dpth " "))) ((list c ct ...) (show-node c (string-append dpth " |")) (l ct)))))
(show-node (tree-root tree) "")</lang>
- Output:
-+ |-- $ |-+ a | |-- $ | `-+ na | |-- $ | `-- na$ |-+ na | |-- $ | `-- na$ `-- banana$
Raku
(formerly Perl 6)
Here is quite a naive algorithm, probably .
The display code is a variant of the visualize a tree task code.
<lang perl6>multi suffix-tree(Str $str) { suffix-tree flat map &flip, [\~] $str.flip.comb } multi suffix-tree(@a) {
hash @a == 0 ?? () !! @a == 1 ?? ( @a[0] => [] ) !! gather for @a.classify(*.substr(0, 1)) { my $subtree = suffix-tree(grep *.chars, map *.substr(1), .value[]); if $subtree == 1 { my $pair = $subtree.pick; take .key ~ $pair.key => $pair.value; } else { take .key => $subtree; } }
}
my $tree = root => suffix-tree 'banana$';
.say for visualize-tree $tree, *.key, *.value.List;
sub visualize-tree($tree, &label, &children,
:$indent = , :@mid = ('├─', '│ '), :@end = ('└─', ' '),
) {
sub visit($node, *@pre) { gather { take @pre[0] ~ $node.&label; my @children = sort $node.&children; my $end = @children.end; for @children.kv -> $_, $child { when $end { take visit($child, (@pre[1] X~ @end)) } default { take visit($child, (@pre[1] X~ @mid)) } } } } flat visit($tree, $indent xx 2);
}</lang>
- Output:
root ├─$ ├─a │ ├─$ │ └─na │ ├─$ │ └─na$ ├─banana$ └─na ├─$ └─na$
Sidef
<lang ruby>func suffix_tree(Str t) {
suffix_tree(^t.len -> map { t.substr(_) })
}
func suffix_tree(a {.len == 1}) {
Hash(a[0] => nil)
}
func suffix_tree(Arr a) {
var h = Hash() for k,v in (a.group_by { .char(0) }) { var subtree = suffix_tree(v.map { .substr(1) }) var subkeys = subtree.keys if (subkeys.len == 1) { var subk = subkeys[0] h{k + subk} = subtree{subk} } else { h{k} = subtree } } return h
}
say suffix_tree('banana$')</lang>
- Output:
Hash( "$" => nil, "a" => Hash( "$" => nil, "na" => Hash( "$" => nil, "na$" => nil ) ), "banana$" => nil, "na" => Hash( "$" => nil, "na$" => nil ) )