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:
import sys
readline = sys.stdin.readline

T = int(readline())

Ans = [None]*T
for qu in range(T):
    N, X = map(int, readline().split())
    table = [0]*(2*N+3)
    cnt = 0
    S = list(map(int, readline().strip()))
    for s in S:
        cnt += (-1 if s else 1)
        table[cnt] += 1
    if cnt == 0:
        table[0] += 1
        if -N <= X <= N and table[X]: 
            Ans[qu] = -1
        else:
            Ans[qu] = 0
        continue
    elif cnt > 0:
        ans = 0 if X else 1
        if N < X:
            k = -((N-X)//cnt)
            X -= k*cnt
        while -N <= X:
            if N < X:
                X -= cnt
                continue
            ans += table[X]
            X -= cnt
    else:
        ans = 0 if X else 1
        if X < -N:
            k = -((-X-N)//cnt)
            X -= k*cnt
        while X <= N:
            if X < -N:
                X -= cnt
                continue
            ans += table[X]
            X -= cnt
    Ans[qu] = ans

print('\n'.join(map(str, Ans)))