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.

from __future__ import division
import sys
input = sys.stdin.readline
import math
from math import sqrt, floor, ceil
from collections import Counter
from copy import deepcopy as dc


############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
def insr2():
    s = input()
    return(s.split(" "))

def prime_factorization(n):

    if n == 1:
        return [1]

    ans=[]
    i = 2
    cap = sqrt(n)
    while i <= cap:
        if n % i == 0:
            ans.append(i)
            n = n//i
            cap=sqrt(n)
        else:
            i += 1
    if n > 1:
        ans.append(n)
    return ans

def binomial(n, k):
    if n == 1 or n == k:
        return 1

    if k > n:
        return 0       
    else:
        a = math.factorial(n)
        b = math.factorial(k)
        c = math.factorial(n-k)
        div = a // (b * c)
        return div 

for _ in range(inp()):

    n, k = invr()
    s = raw_input()

    if n%2 == 0 and k == n//2:
        print 'NO'
        continue

    f = True

    for i in range(k):
        if s[i] != s[len(s)-1-i]:
            f == False
            print 'NO'
            break
    else:
        print 'YES'
            