t=int(input())
while t:
n,q=map(int,input().split())
s=input()
for i in range(q):
l,r=map(int,input().split())
l-=1
r-=1
if (s[l] in s[:l]) or (s[r] in s[r+1:]):
print('YES')
else:
print('NO')
t-=1#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <ll> vll;
typedef vector <char> vc;
typedef vector <string> vs;
typedef vector <pair<int, int>> vpii;
typedef vector <pair<char, int>> vpci;
typedef pair <int, int> pii;
typedef map <int, int> mii;
typedef map <char, int> mci;
typedef map <string, int> msi;
typedef set <int> si;
typedef set <char> sc;
typedef set <string> ss;
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define all(c) (c).begin(), (c).end()
#define lbnd lower_bound
#define ubnd upper_bound
#define debug cerr<<"\ni'm here\n"<<endl;
#define PI 3.141592653589793
const ll MAX_N = 1000000007;
// resolvendo usando guloso:
// so sera possivel nos dois seguintes casos:
// (1) tem char igual a str[l-1] antes de str[l-1]
// (2) tem char igual a str[r-1] depois de str[r-1]
void solve(){
int n, q, l, r;
string str;
cin >> n >> q >> str;
for(int i=0; i<q; i++){
cin >> l >> r;
bool poss = false;
for(int k=0; k<(l-1); k++){
if(str[k]==str[l-1]){
poss = true;
}
}
for(int j=n-1; j>(r-1); j--){
if(str[j]==str[r-1]){
poss = true;
}
}
if(!poss)
cout << "NO" << endl;
else
cout << "YES" << endl;
}
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
while(t--)
solve();
//solve();
return 0;
} 1302. Deepest Leaves Sum | 1209. Remove All Adjacent Duplicates in String II |
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