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 pb push_back
#define mod 10000007
#define ll long long int
#define INF 99999999
// int bit[1000001];
// //for adding x to required positions in bit array
// void add(int i,int x)
// {
//   for(;i<1000001;i+=(i&(-i)))
//   {
//      bit[i]+=x;
//   }
// }
// //for finding sum from 0 to i inclusive
// int find(int i)
// {
//   int sum=0;
//   for(;i>0;i-=(i&(-i)))
//   {
//        sum+=bit[i];
//   }
//   return sum;
// }


int main()
{ 
 int t;
 cin>>t;
 while(t--)
 {
	 int n,k;
	 cin>>n>>k;
	 string s;
	 cin>>s;
	 if(n<=2*k)
	 {
		 cout<<"NO"<<endl;
	 }
	 else
	 {
		 string one="";
		 string two="";
		 for(int i=0;i<k;i++)
		 {
			 one+=s[i];
		 }
		 for(int i=n-1;i>=n-k;i--)
		 {
			 two+=s[i];
		 }
		
		

		 if(one==two)
		 {
			 cout<<"YES"<<endl;
		 }
		 else
		 {
			 cout<<"NO"<<endl;
		 }
	 }
 }
}