#include <bits/stdc++.h>

#define endl '\n'
#define fi first
#define se second
#define forn(i,n) for (int i = 0; i < n; i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb push_back
#define fastIO(); ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define SZ(s) int(s.size())
#define var(x) #x << " = " << x

using namespace std;

typedef long long lli;
typedef long double ld;
typedef pair<lli, lli> plli;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<vector<lli>> vvlli;
typedef vector<lli> vlli;
typedef vector<pii> vpii;

int main () {
	fastIO();
	int t = 1;
	cin >> t;
	while(t--){
		string s;
		int n, k;
		cin >> n >> k >> s;
		int l, r;
		l = 0;
		r = n - 1;
		while((r - l) > 1){
			if(s[l] == s[r]){
				l++;
				r--;
			}
			else
				break;
		}
		if(l >= k) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>

#define endl '\n'
#define fi first
#define se second
#define forn(i,n) for (int i = 0; i < n; i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb push_back
#define fastIO(); ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define SZ(s) int(s.size())
#define var(x) #x << " = " << x

using namespace std;

typedef long long lli;
typedef long double ld;
typedef pair<lli, lli> plli;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<vector<lli>> vvlli;
typedef vector<lli> vlli;
typedef vector<pii> vpii;

int main () {
	fastIO();
	int t = 1;
	cin >> t;
	while(t--){
		string s;
		int n, k;
		cin >> n >> k >> s;
		int l, r;
		l = 0;
		r = n - 1;
		while((r - l) > 1){
			if(s[l] == s[r]){
				l++;
				r--;
			}
			else
				break;
		}
		if(l >= k) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
	return 0;
}

```