F. Isomorphic Strings

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 T for which f(si) = ti. Formally:

1. f(si) = ti 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: xi, yi and leni (1 ≤ xi, yi ≤ n, 1 ≤ leni ≤ n - max(xi, yi) + 1) — the description of the pair of the substrings to check.

Output

For each query in a separate line print "YES" if substrings s[xi... xi + leni - 1] and s[yi... yi + leni - 1] are isomorphic and "NO" otherwise.

Example

input

Copy

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

output

Copy

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.

題意：給你一個字串，問你裡面兩個長度相同的子串，裡面的字母能否一一對應

解題思路：把一個字串拆成26個01字串，如abcdbc變為

100000

010010

001001

000100

...

然後對這些字串做雜湊，每次查詢每個字母的子串的hash值是否相等即可。

這裡用來儲存字串雜湊模板

#include <iostream>
#include <vector>
#include <string.h>
#include <algorithm>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int MAXN = 200005;

struct HashTable
{
    ull g[MAXN];
    ull P1 = 1997;
    ull P2 = 10000007;
    ull Hash[MAXN];
    ull getHash(int x, int len)
    {
        return ((Hash[x + len - 1] - Hash[x - 1] * g[len] % P2) + P2) % P2;
    }
    void hash(char *s, int len)
    {   
        g[0] = 1;
        for (int i = 1; i <= MAXN-1; i++)
        g[i] = (g[i - 1]) * P1 % P2;
        for (int i = 1; i <= len; i++)
            Hash[i] = (Hash[i - 1] * P1 + s[i - 1]) % P2;
    }
};
HashTable table[26];
char s[MAXN];
char a[26][MAXN];
int ans1[MAXN];
int ans2[MAXN];
int main()
{
    
    int N, Q;
    scanf("%d%d", &N, &Q);
    scanf("%s", s);
    int len = strlen(s);
    for (int i = 0; i < len; i++)
    {
        for (int j = 0; j < 26; j++)
        {
            a[j][i] = ((s[i] - 'a') == j) + 1;
        }
    }
    for (int i = 0; i < 26; i++)
    {
        table[i].hash(a[i], len);
    }

    int x, y, l;
    while (Q--)
    {
        scanf("%d%d%d", &x, &y, &l);

        for (int i = 0; i < 26; i++)
        {
            ans1[i] = table[i].getHash(x, l);
            ans2[i] = table[i].getHash(y, l);
        }
        sort(ans1, ans1 + 26);
        sort(ans2, ans2 + 26);
        bool flag = 0;
        for (int i = 0; i < 26; i++)
        {
            if (ans1[i] != ans2[i])
            {
                flag = 1;
                cout << "NO" << endl;
                break;
            }
        }
        if (flag == 0)
            cout << "YES" << endl;
    }

    return 0;
}