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:
for _ in range(int(input())):
    n, x = map(int, input().split())
    s, ans, d = input(), 0, {}
    for q in s:
        d[ans], ans = d.get(ans, 0)+1, ans+(q == '0')*2-1
    if ans == 0:
        print(-(x in d))
    else:
        if ans > 0:
            x, answer, q1 = x-(x-100001)//ans*ans*(x > 100001), 0, 0
        else:
            x, answer, q1 = x+(-x-100001)//(-ans)*(-ans) * (x < -100001), 0, 0
        while (x-q1 > -100001) if ans > 0 else x-q1 < 100001:
            if x - q1 in d:
                answer += d[x - q1]
            q1 += ans
        print(answer)