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.
///******* In the name of Allah *******///
//  AL-HAMDULILLAH
//  AL-HAMDULILLAH
//  AL-HAMDULILLAH
 
//#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define ll long long
#define size 1000001
#define fopenr() freopen("input.txt", "r", stdin)
#define fopenw() freopen("output.txt", "w", stdout)
#define FastIO ios_base::sync_with_stdio(false), cin.tie(0)
#define bm 1000000007   //bigmod
#define reset(arr,n,i) fill(arr,arr+n,i)  //cbrt(n) means cube root
using namespace std;
 
void solve()
{
	int n,i=0,k,c=0,count=0;
	cin>>n>>k;
	string str;
	cin>>str;
	
	for(i=0;i<n/2;i++)
	{
		if(str[i]==str[n-i-1] && count!=k)
			count++;
		else
			break;
	}
	
	if(count==k && k<<1<n)
		cout<<"YES"<<endl;
	else
		cout<<"NO"<<endl;
}
 
int main()
{
	int n;
	cin>>n;
	while(n--)
	{
		solve();
	}
	return 0;
}
//written by mamun01(nameoa01)