Non-Substring Subsequence CodeForces - 1451B（字符串处理问题）

Hr0d1y has q queries on a binary string s of length n. A binary string is a string containing only characters ‘0’ and ‘1’.

A query is described by a pair of integers li, ri (1≤li<ri≤n).

For each query, he has to determine whether there exists a good subsequence in s that is equal to the substring s[li…ri].

A substring s[i…j] of a string s is the string formed by characters sisi+1…sj.

String ais said to be a subsequence of string b if a can be obtained from b by deleting some characters without changing the order of the remaining characters.

A subsequence is said to be good if it is not contiguous and has length ≥2. For example, if s is “1100110”, then the subsequences s1s2s4 (“1100110”) and s1s5s7 (“1100110”) are good, while s1s2s3 (“1100110”) is not good.

Can you help Hr0d1y answer each query?

Input

The first line of the input contains a single integer t (1≤t≤100 ) — the number of test cases. The description of each test case is as follows.

The first line contains two integers n(2≤n≤100) and q (1≤q≤100) — the length of the string and the number of queries.

The second line contains the string s.

The i-th of the next q lines contains two integers li and ri (1≤li<ri≤n ).

Output

For each test case, output q lines. The i-th line of the output of each test case should contain “YES” if there exists a good subsequence equal to the substring s[li…ri] , and “NO” otherwise.

You may print each letter in any case (upper or lower).

Example

Input

2
6 3
001000
2 4
1 3
3 5
4 2
1111
1 4
2 3

Output

YES
NO
YES
NO
YES

Note

In the first test case, s[2…4]=“010”. In this case s1s3s5 (“001000”) and s2s3s6 (“001000”) are good suitable subsequences, while s2s3s4
(“001000”) is not good.
s[1…3]=“001”. No suitable good subsequence exists.
s[3…5]=“100”. Here s3s5s6 (“001000”) is a suitable good
subsequence.

题意：

先给出一个区间，在区间内的字符串组成一个新的字符串，求原来的字符串除给出的区间外还有没有与新的字符串相同的字符串（下标不可以连续）有输出Yes，否则No

思路：

中间不动，对比两边
判断是否还有相同的字符串，可以先判断区间外前面的范围是否有和区间内第一个字符相同的，如果有，就可以实现下标不连续（意思就是将区间的第一个字符的下标换为前面与其相同的字符的下标，同理，对比区间后面的也是如此）
一句话思路：如果区间前面或后面能找到与新字符串的第一个字符或最后一个字符相同的，就输出Yes

代码：

#include<stdio.h>
char a[120];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,q,l,r;
        scanf("%d %d",&n,&q);
        scanf("%s",a+1);
        while(q--)
        {
            int j=0;
            scanf("%d %d",&l,&r);
            int f=0,f1=0;
            for(int i=1; i<l; i++)
            {
                if(a[i]==a[l])
                    f=1;
            }
            for(int i=r+1; i<=n; i++)
            {
                if(a[i]==a[r])
                    f1=1;
            }
            if(f==1||f1==1)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}