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:
def n_matches(s, x):
    bal_s = len(s)-2*sum(s)   # n1 is sum(s), n0 is len(s)-sum(s) --> n0-n1 = len(s)-2*sum(s)
    run_bal = 0
    # case bal_s == 0
    if bal_s == 0:
        for s_i in s:
            if s_i == 0:
                run_bal += 1
            else:
                run_bal -= 1
            if run_bal == x:
                return -1 # if x is matched once, it will be matched inf times
        return 0 # if x is not matched inside one sequence it will never be matched
    # case bal_s != 0
    n_matches = 0
    if x == 0:
        n_matches += 1
    for s_i in s:
        if s_i == 0:
            run_bal += 1
        else:
            run_bal -= 1
        if (x-run_bal)//bal_s >= 0 and (run_bal-x)%bal_s == 0:
            n_matches += 1
    return n_matches

def test_input():
    with open(r"C:\Users\Enrico\Desktop\CodeForces\1295\B\input.txt") as f:
        T = int(f.readline())
        S = [0]*T
        X = [0]*T
        for i in range(T):
            [n, X[i]] = [int(i) for i in f.readline().split()]
            S[i] = [int(i) for i in f.readline().strip()]
    return T, S, X

def real_input():
    T = int(input())
    S = [0]*T
    X = [0]*T
    for i in range(T):
        [n, X[i]] = [int(i) for i in input().split()]
        S[i] = [int(i) for i in input().strip()]
    return T, S, X

T, S, X = real_input()
matches = [0]*T
for i in range(T):
    matches[i] = str(n_matches(S[i], X[i]))
out = "\n".join(matches)
print(out)