CodeForces - 985F Isomorphic Strings

假如两个区间的26的字母出现的位置集合分别是 A1,B1,A2,B2,....., 我们再能找到一个排列p[] 使得 A[i] = B[p[i]] ,那么就可以成功映射了。

显然集合可以直接hash，排序一下hash值就可以判断是否匹配了。。。。

虽然不知道为什么要卡19260817，但反正我是坚决不写双hash的，最后随便换了一个大模数(甚至都不是质数2333)然后就过了。。。

Discription

You are given a string s of length n consisting of lowercase English letters.

For two given strings s and t, say S is the set of distinct characters of s and T is the set of distinct characters of t. The strings s and t are isomorphic if their lengths are equal and there is a one-to-one mapping (bijection) f between S and Tfor which f(s_i) = t_i. Formally:

1. f(s_i) = t_i for any index i,
2. for any character  there is exactly one character  that f(x) = y,
3. for any character  there is exactly one character  that f(x) = y.

For example, the strings "aababc" and "bbcbcz" are isomorphic. Also the strings "aaaww" and "wwwaa" are isomorphic. The following pairs of strings are not isomorphic: "aab" and "bbb", "test" and "best".

You have to handle m queries characterized by three integers x, y, len (1 ≤ x, y ≤ n - len + 1). For each query check if two substrings s[x... x + len - 1] and s[y... y + len - 1] are isomorphic.

Input

The first line contains two space-separated integers n and m (1 ≤ n ≤ 2·10⁵, 1 ≤ m ≤ 2·10⁵) — the length of the string s and the number of queries.

The second line contains string s consisting of n lowercase English letters.

The following m lines contain a single query on each line: x_i, y_i and len_i (1 ≤ x_i, y_i ≤ n, 1 ≤ len_i ≤ n - max(x_i, y_i) + 1) — the description of the pair of the substrings to check.

Output

For each query in a separate line print "YES" if substrings s[x_i... x_i + len_i - 1] ands[y_i... y_i + len_i - 1] are isomorphic and "NO" otherwise.

Example

Input

7 4
abacaba
1 1 1
1 4 2
2 1 3
2 4 3

Output

YES
YES
NO
YES

Note

The queries in the example are following:

1. substrings "a" and "a" are isomorphic: f(a) = a;
2. substrings "ab" and "ca" are isomorphic: f(a) = c, f(b) = a;
3. substrings "bac" and "aba" are not isomorphic since f(b) and f(c) must be equal to a at same time;
4. substrings "bac" and "cab" are isomorphic: f(b) = c, f(a) = a, f(c) = b.

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=200005,ha=998244357,b=233;
inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;}
int H[maxn][26],ci[maxn],A[26],B[26];
int n,Q,X,Y,L;
char ch;

inline void Get(int *x,int y){
	for(int i=0;i<26;i++) x[i]=add(H[y+L-1][i],ha-H[y-1][i]*(ll)ci[L]%ha);
}

inline void solve(){
	int i;
	while(Q--){
		scanf("%d%d%d",&X,&Y,&L);
		Get(A,X),Get(B,Y);

		sort(A,A+26),sort(B,B+26);

		for(i=0;i<26;i++) if(A[i]!=B[i]) break;
		puts(i==26?"YES":"NO");
	}
}

int main(){
	scanf("%d%d",&n,&Q);

	ci[0]=1;
	for(int i=1;i<=n;i++) ci[i]=(ci[i-1]*(ll)b)%ha;

	for(int i=1;i<=n;i++){
		ch=getchar();
		while(ch<'a'||ch>'z') ch=getchar();

		for(int j=0;j<26;j++) H[i][j]=(H[i-1][j]*(ll)b)%ha;
		H[i][ch-'a']++;
	}

	solve();

	return 0;
}