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:
from collections import defaultdict


def balance_is_realizable(imbalance, x, balance):
    if imbalance >= 0:
        return (x - balance) >= 0
    else:
        return (x - balance) <= 0


def main():
    t = int(input())
    for i in range(t):
        n, x = tuple(int(o) for o in input().split())
        string = input()
        balances_to_occurrences = defaultdict(int)
        imbalance = 0
        s = 0
        for char in string:
            if char == '0':
                imbalance += 1
            else:
                imbalance -= 1
            balances_to_occurrences[imbalance] += 1
        if imbalance == 0:
            if (x in balances_to_occurrences or x == 0):
                print(-1)
            else:
                print(0)
        else:
            for balance, occurrences in balances_to_occurrences.items():
                if (x-balance)%imbalance == 0 and balance_is_realizable(imbalance, x, balance):
                    s += occurrences
            if x== 0:
                s += 1
            print(s)


if __name__ == "__main__":
    main()