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.
#include "bits/stdc++.h"
using namespace std;

using ll = long long;

using vi = vector<int>;
#define pb push_back
#define rsz resize
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()

using pi = pair<int,int>;
#define f first
#define s second
#define mp make_pair

const int mxn = 1e2, mxa = 1e6;

int n, k;
string s;

void solve() {
    cin >> n >> k >> s;
    
    if (k == 0) {
        cout << "YES\n";
        return;
    }

    for (int i = k; 2*i < n; ++i) {
        string s1 = s.substr(0, i), s2 = s.substr(n-i, i);
        reverse(s2.begin(), s2.end());
        if (s1 == s2) {
            cout << "YES\n";
            return;
        }
    }

    cout << "NO\n";
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);

    int t;
    // t = 1;
    cin >> t;
    
    while (t --> 0) solve();
}
