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.
///...................SUBHASHIS MOLLICK...................///
///.....DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING....///
///.............ISLAMIC UNIVERSITY,BANGLADESH.............///
///....................SESSION-(14-15)....................///
#include<bits/stdc++.h>
using namespace std;
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define sf(a) scanf("%d",&a)
#define sf2(a,b) scanf("%d %d",&a,&b)
#define sf3(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define pf(a) printf("%d",a)
#define pf2(a,b) printf("%d %d",a,b)
#define pf3(a,b,c) printf("%d %d %d",a,b,c)
#define sfl(a) scanf("%lld",&a)
#define sfl2(a,b) scanf("%lld %lld",&a,&b)
#define sfl3(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define pfl(a) printf("%lld",a)
#define pfl2(a,b) printf("%lld %lld",a,b)
#define pfl3(a,b,c) printf("%lld %lld %lld",a,b,c)
#define nl printf("\n")
#define   timesave              ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ll long long
#define pb push_back
#define MPI map<int,int>mp;
#define fr(i,n) for(i=0;i<n;i++)
#define fr1(i,n) for(i=1;i<=n;i++)
#define frl(i,a,b) for(i=a;i<=b;i++)
#define tz 100000
#define clr0(a) memset(a,0,sizeof(a))
#define clr1(a) memset(a,-1,sizeof(a))
#define space " "
#define yesp cout<<"YES"<<endl;
#define nop cout<<"NO"<<endl;
/*primes in range 1 - n
1 - 100(1e2) -> 25 pimes
1 - 1000(1e3) -> 168 primes
1 - 10000(1e4) -> 1229 primes
1 - 100000(1e5) -> 9592 primes
1 - 1000000(1e6) -> 78498 primes
1 - 10000000(1e7) -> 664579 primes
large primes ->
104729 1299709 15485863 179424673 2147483647 32416190071 112272535095293 48112959837082048697
*/
//freopen("Input.txt","r",stdin);
//freopen("Output.txt","w",stdout);
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
int main()
{
    timesave;
    //freopen("Input.txt","r",stdin);
    //freopen("Output.txt","w",stdout);
    int ts=1;
    cin>>ts;
    while(ts--)
    {
        ll n,k;
        string s;
        cin>>n>>k>>s;
        int flag=0;
        for(int i=0,j=n-1; i<k; i++,j--)
        {
            if(s[i]!=s[j])
                flag=1;
            if(i+1==j)
                flag=1;
        }
        if(flag)
            cout<<"NO"<<endl;
        else
            cout<<"YES"<<endl;

    }
}
