Kawashiro Nitori is a girl who loves competitive programming.

One day she found a string and an integer. As an advanced problem setter, she quickly thought of a problem.

Given a string s and a parameter k, you need to check if there exist k+1 non-empty strings a_1,a_2...,a_{k+1}, such that $$$s=a_1+a_2+… +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+…+R(a_{1}).$$$ 

Here + represents concatenation. We define R(x) as a reversed string x. For example R(abcd) = dcba. Note that in the formula above the part R(a_{k+1}) is intentionally skipped.

Input

The input consists of multiple test cases. The first line contains a single integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case description contains two integers n, k (1≤ n≤ 100, 0≤ k≤ ⌊ n/2 ⌋) — the length of the string s and the parameter k.

The second line of each test case description contains a single string s of length n, consisting of lowercase English letters.

Output

For each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,…,a_{k+1}, and "NO" (without quotes) otherwise.

You can print letters in any case (upper or lower).

Example

Input


7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa


Output


YES
NO
YES
NO
YES
NO
NO

Note

In the first test case, one possible solution is a_1=qw and a_2=q.

In the third test case, one possible solution is a_1=i and a_2=o.

In the fifth test case, one possible solution is a_1=dokidokiliteratureclub.
 
//     THE ONLY WAY IS UP
 
#include<bits/stdc++.h>
 
#define nl              "\n"

using namespace std;

bool check(string s, int n) {

    for(int i=0; i<ceil(n/2); i++) {
        if(s[i] == s[n-i-1]) { continue; }
        return false;
    }
    return true;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int tt; cin>>tt;
    while(tt--) {
        int n; cin>>n;
        int k; cin>>k;
        string s; cin>>s;
        
        if(k == 0) { cout<<"YES"<<nl; continue; }

        int i=0;
        for(; i<ceil(n/2); i++) {
            if(s[i] == s[n-i-1]) { continue; }
            else break;
        }

        if(n%2 == 0) { i--; }
        if(k <= i) { cout<<"YES"<<nl; }
        else cout<<"NO"<<nl;




//        if(check(s, n)){ 
//            if(n >= k*2 + 1) { cout<<"YES"<<nl; }
//            else { cout<<"NO"<<nl; }
//        }else {
//            cout<<"NO"<<nl;
//        }
    }
}
