A. Split it!time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputKawashiro 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+\ldots +a_k+a_{k+1}+R(a_k)+R(a_{k-1})+\ldots+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.InputThe input consists of multiple test cases. The first line contains a single integer t (1\le t\le 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\le n\le 100, 0\le k\le \lfloor \frac{n}{2} \rfloor)  — 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.OutputFor each test case, print "YES" (without quotes), if it is possible to find a_1,a_2,\ldots,a_{k+1}, and "NO" (without quotes) otherwise.You can print letters in any case (upper or lower).ExampleInput
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
NoteIn 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.
7
5 1
qwqwq
2 1
ab
3 1
ioi
4 2
icpc
22 0
dokidokiliteratureclub
19 8
imteamshanghaialice
6 3
aaaaaa

YES NO YES NO YES NO NO 
['brute force', 'constructive algorithms', 'greedy', 'strings', '*900']