Solve the programming task below in a Python markdown code block.
Real stupidity beats artificial intelligence every time.

— Terry Pratchett, Hogfather, Discworld

You are given a string $s$ of length $n$ and a number $k$. Let's denote by $rev(s)$ the reversed string $s$ (i.e. $rev(s) = s_n s_{n-1} ... s_1$). You can apply one of the two kinds of operations to the string:

replace the string $s$ with $s + rev(s)$

replace the string $s$ with $rev(s) + s$

How many different strings can you get as a result of performing exactly $k$ operations (possibly of different kinds) on the original string $s$?

In this statement we denoted the concatenation of strings $s$ and $t$ as $s + t$. In other words, $s + t = s_1 s_2 ... s_n t_1 t_2 ... t_m$, where $n$ and $m$ are the lengths of strings $s$ and $t$ respectively.


-----Input-----

The first line contains one integer $t$ ($1 \le t \le 100$) — number of test cases. Next $2 \cdot t$ lines contain $t$ test cases:

The first line of a test case contains two integers $n$ and $k$ ($1 \le n \le 100$, $0 \le k \le 1000$) — the length of the string and the number of operations respectively.

The second string of a test case contains one string $s$ of length $n$ consisting of lowercase Latin letters.


-----Output-----

For each test case, print the answer (that is, the number of different strings that you can get after exactly $k$ operations) on a separate line.

It can be shown that the answer does not exceed $10^9$ under the given constraints.


-----Examples-----

Input
4
3 2
aab
3 3
aab
7 1
abacaba
2 0
ab
Output
2
2
1
1


-----Note-----

In the first test case of the example:

After the first operation the string $s$ can become either aabbaa or baaaab. After the second operation there are 2 possibilities for $s$: aabbaaaabbaa and baaaabbaaaab.