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:
"""
Author: guiferviz
Time: 2020-01-29 20:26:47

Example:
 n=6  x=10
 s=010010
 Positions where q matches x are between ().
 010010 010010 010010 010010 010(0)1(0) 0(1)0010
"""

def solve():
    # Read input.
    n, x = map(int, input().split())
    s = input()

    # Precompute balance q in a list.
    q = [0]
    for i, s_i in enumerate(s):
        q.append(q[i] + (1 if s_i == "0" else -1))
    q = q[1:]

    sol = 1 if x == 0 else 0
    for i in q:
        # Infinite case: after one cycle the balance is 0.
        if q[-1] == 0:
            if i == x:
                sol = -1
        # Finite case: q is increasing or decreasing, so it's going to pass x.
        else:
            need = x - i
            if need % q[-1] == 0 and need * q[-1] >= 0:
                # q is going to match x in iteration `need // q[-1]` and
                # position inside sequence given by this for loop.
                sol += 1
    print(sol)


def main():
    t = int(input())
    for _ in range(t):
        solve()


if __name__ == "__main__":
    main()