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;
#define dd double
#define ll long long int
#define light ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define t_pair pair<ll,pair<ll,ll> >
#define pb push_back
ll INF = 1e9 + 5;
ll mod = 998244353;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        int k;
        cin >> n >> k;
        string s;
        cin >> s;
        int i = 0;
        int j = s.length() - 1;
        if (k == 0)
        {
            cout << "YES\n";
            continue;
        }
        int count = 0;
        while (j > i)
        {
            if (s[i] == s[j])
            {
                count++;
            }
            else
            {
                break;
            }
            i++;
            j--;

        }
        if (j >= i)
        {
            if (count >= k)
            {
                cout << "YES\n";
            }
            else
            {
                cout << "NO\n";
            }
        }
        if (j < i)
        {
            count--;
            if (count >= k)
            {
                cout << "YES\n";
            }
            else
            {
                cout << "NO\n";
            }
        }
    }

}