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 ull = unsigned long long;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using mi = map<int, int>;
using ml = map<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;

#define FOR(i, a, b) for(int i = a; i < b; ++i)
#define F0R(i, b) FOR(i, 0, b)
#define ROF(i, b, a) for(int i = b; i >= a; --i)
#define R0F(i, b) ROF(i, b, 0)
#define trav(x, a) for(auto &x: a)

#define mp make_pair
#define mtp make_tuple
#define pb push_back
#define pf push_front

#define sz(x) int(x.size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

void solve()
{
	int n, k; cin >> n >>k;
    string s; cin >> s;
    if(k == 0)
    {
        cout << "YES\n";
        return;
    }
    if(k*2 == n)
    {
        cout << "NO\n";
        return;
    }

    F0R(i, k)
    {
        if(s[i] != s[sz(s)-1-i])
        {
            cout << "NO\n";
            return;
        }
    }
    cout << "YES\n";
}

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);

	int t;
	cin >> t;
	while(t--)
	{
		solve();
	}
	return 0;
}