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.
import java.util.*;
import java.io.*;

public class A{
	private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	private static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
	
	public static void main (String[] args) throws IOException {
        int t = Integer.parseInt(br.readLine());
        while(t-- > 0) {
            String nk[] = br.readLine().split(" ");
            int n = Integer.parseInt(nk[0]);
            int k = Integer.parseInt(nk[1]);
            String s = br.readLine();
            if(n < (2*k)+1) {
                bw.write("NO\n");
                continue;
            }
            int f = 0, l = n-1;
            int count = 0;
            String a = "", b = "";
            boolean ans = false;
            while(f<l) {
                if(count == k) {
                    if(l-f >= 0) {
                        ans = true;
                    }
                    break;
                }
                a += Character.toString(s.charAt(f));
                b += Character.toString(s.charAt(l));
                if(a.equals(b)) {
                    count++;
                    a = "";
                    b = "";
                }
                f++;
                l--;
            }
            if(count == k && l-f>=0) {
                ans = true;
            }
            if(ans) {
                bw.write("YES\n");
            }else {
                bw.write("NO\n");
            }
        }
        bw.flush();
    }
}