#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
const int N = 200000;
const int M = 1000000007;

int n, k;
string s;


void solve() {
  int good = 0;
  for (int i = 0 ; i < n >> 1 ; i++) {
    if (s[i] == s[n - i - 1])
      good++;
    else
      break;
  }
  cout << (good >= k && k * 2 < n ? "YES" : "NO") << endl;


}

void read() {
  cin >> n >> k >> s;

}

signed main() {
  int t = 1;
  cin >> t;
  while (t--) {
    read();
    solve();
  }
}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
const int N = 200000;
const int M = 1000000007;

int n, k;
string s;


void solve() {
  int good = 0;
  for (int i = 0 ; i < n >> 1 ; i++) {
    if (s[i] == s[n - i - 1])
      good++;
    else
      break;
  }
  cout << (good >= k && k * 2 < n ? "YES" : "NO") << endl;


}

void read() {
  cin >> n >> k >> s;

}

signed main() {
  int t = 1;
  cin >> t;
  while (t--) {
    read();
    solve();
  }
}

```