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 l_i, r_i (1 ≤ l_i < r_i ≤ n). 

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

  * A substring s[i… j] of a string s is the string formed by characters s_i s_{i+1} … s_j.
  * String a is 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 s_1s_2s_4 ("1100110") and s_1s_5s_7 ("1100110") are good, while s_1s_2s_3 ("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 l_i and r_i (1 ≤ l_i < r_i ≤ 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[l_i...r_i], 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 s_1s_3s_5 ("001000") and s_2s_3s_6 ("001000") are good suitable subsequences, while s_2s_3s_4 ("001000") is not good. 
  * s[1… 3] =  "001". No suitable good subsequence exists. 
  * s[3… 5] =  "100". Here s_3s_5s_6 ("001000") is a suitable good subsequence. 
#include<bits/stdc++.h>
using namespace std;
 #define MAX 1000000000005
 #define ll long long int
           vector<ll>a[1000],w;
            map<ll,vector<ll>>mp;
            ll b[1000]={0},c=0;
 ll e=1e7+5;
   vector<ll>p(e+10000,1),g;
   void sieve(){
     ll i,j;
           p[0]=0;
        p[1]=0;
        for(int i=2;i*i<e;i++)
        {if(p[i])
        { for(j=i*i;j<e;j+=i)
            p[j]=0;
        }
        }
   }
    vector<ll>fact(ll x)
    {  vector<ll>v;
     ll i;
        if(x%2==0)
    {   v.push_back(2);
         while(x%2==0)
            x/=2;
    }
     for(i=3;i<=sqrt(x);i+=2)
     { if(x%i==0)
        {    v.push_back(i);
            while(x%i==0)
            x/=i;
        }
     }
     if(x>2)
         v.push_back(x);
          return v;
    }
    void fact()
    {  ll i,j;
     for(i=1;i<=100000;i++)
     {for(j=1;j<=sqrt(i);j++)
    if(i%j==0)
         {mp[i].push_back(j);
           if(i/j!=j)
            mp[i].push_back(i/j);
         }}}
         void dfs(ll d)
         {  b[d]=1;
            for(auto x:a[d])
            { if(b[x]==0)
                 dfs(x);
            }
         }
  int main()
 {    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
         ll t,i,j;
           cin>>t;
            while(t--){
              ll n,x=0,q,y,d;
               vector<ll>v;
                 map<ll,ll>mp;
                 cin>>n>>q;
                  string s;
                   cin>>s;
                     while(q--)
                     { cin>>x>>y;
                       x--;
                       y--;
                      c=0;d=0;
                       for(i=0;i<x;i++)
                       { if(s[i]==s[x])
                           c=1;
                       }
                       for(i=y+1;i<n;i++)
                       { if(s[i]==s[y])
                           d=1;
                       }
                       if(c==1||d==1)
                          cout<<"YES"<<endl;
                       else
                        cout<<"NO"<<endl;
                     }

                     }
 }
