#include <algorithm>
#include <iostream>
#include <iomanip>
#include <vector>
#include <cmath>
#include <map>
#include <set>

#define all(a) a.begin(), a.end()
#define pb push_back

using namespace std;
using ll = long long;
using ld = long double;

const int N = 2e5+7;

void solve() {
    int n, k; cin >> n >> k;
    string s; cin >> s;
    string t1, t2;
    for(int i = 0; i < k; ++i) t1 += s[i];
    for(int i = n - k; i < n; ++i) t2 += s[i];
    reverse(all(t1));
    if(t1 == t2 && 2*k < n) cout << "YES\n";
    else cout << "NO\n";
}

int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    int t; cin >> t;
    while(t--) solve();
    return 0;
}

### 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 <algorithm>
#include <iostream>
#include <iomanip>
#include <vector>
#include <cmath>
#include <map>
#include <set>

#define all(a) a.begin(), a.end()
#define pb push_back

using namespace std;
using ll = long long;
using ld = long double;

const int N = 2e5+7;

void solve() {
    int n, k; cin >> n >> k;
    string s; cin >> s;
    string t1, t2;
    for(int i = 0; i < k; ++i) t1 += s[i];
    for(int i = n - k; i < n; ++i) t2 += s[i];
    reverse(all(t1));
    if(t1 == t2 && 2*k < n) cout << "YES\n";
    else cout << "NO\n";
}

int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    int t; cin >> t;
    while(t--) solve();
    return 0;
}

```