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;
using ll = long long;
using llu = unsigned long long;
using ld = long double;

#define CEIL(x, y) ceil((ld)x / (ld)y)

#define dbgLST(ds) for(auto it : ds) cout << it << ' ';
#define dbgMP(ds) for(auto it : ds) cout << it.first << ' ' << it.second << '\n';

const double PI = acos(-1.0), g = 9.80665;
const ll unseen = 0, pinf = 1e18 + 505, ninf = -1e18 - 505, md = 1000000007;
ll i, j, k, n, m;
bool is;  

void solve(void)
{
    cin >> n >> k;
    string s;
    cin >> s;
    if(k == 0) cout << "YES\n";
    else if(n % 2 == 0 && k == (n / 2)) cout << "NO\n";
    else {
        bool isp = 1;
        ll sz;
        for(i = 0, j = s.size() - 1; i < k; i++, j--) {
            isp &= (s[i] == s[j]);
        }
        cout << (isp ? "YES\n" : "NO\n");
    }
}

int main()
{
    ios_base::sync_with_stdio(false); 
    cin.tie(nullptr); 

    int testcase = 1;
    cin >> testcase;
    while(testcase--) {
        solve();
    }    
    return 0;
}