Problem: You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...

Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.

A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next 2T lines contain descriptions of test cases — two lines per test case. The first line contains two integers n and x (1 ≤ n ≤ 10^5, -10^9 ≤ x ≤ 10^9) — the length of string s and the desired balance, respectively.

The second line contains the binary string s (|s| = n, s_i ∈ \{0, 1\}).

It's guaranteed that the total sum of n doesn't exceed 10^5.

Output

Print T integers — one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.

Example

Input


4
6 10
010010
5 3
10101
1 0
0
2 0
01


Output


3
0
1
-1

Note

In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.

Solution (in PYTHON3):

Solution:
class Solution(object):
    def __init__(self):
        pass

    def apply(self, s, x):

        b = 0
        prefix = []
        for c in s:
            if c == '1':
                b -= 1
            else:
                b += 1
            prefix.append(b)

        
        s = 0
        for nm in prefix:
            if b == 0:
                if nm == x:
                    return -1
            else:
                if (x - nm) % b == 0 and (x - nm) * b >= 0:
                    s += 1
        if x == 0:
            s += 1
        return s


def main():
    sol = Solution()
    t = int(input())
    for i in range(t):
        info = input().split()
        n = int(info[0])
        x = int(info[1])
        s = input()
        print(sol.apply(s, x))

    return 0

if __name__ == "__main__":
    main()