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 mod 1000000007
#define pie  3.14159265
#define ll long long int
#define fr(i,a,n) for(ll i=a;i<n;i++)
#define vi vector<int>
#define vlli vector<ll>
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define fv(i,a) for(auto i:a)
#define sz(a) int((a).size())
#define all(c) (c).begin(),(c).end()
#define frv(i,c) for(auto i = c.begin(); i != c.end(); i++)

void solve()
{
	ll n, k;
	cin >> n >> k;
	string s;
	cin >> s;
	if (n >= ((2 * k) + 1))
	{
		fr(i, 0, k)
		{
			if (s[i] != s[n - 1 - i])
			{
				cout << "NO" << endl; return;
			}
		}
		cout << "YES" << endl; return;
	}
	cout << "NO" << endl; return;
}


int main() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen ("output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t;
	cin >> t;
	//t = 1;
	while (t--)
	{
		solve();
	}


	return 0;
}