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 PYTHON):

Solution:
from __future__ import division
from sys import stdin

rstr = lambda: stdin.readline().strip()
rints = lambda: [int(x) for x in stdin.readline().split()]

for _ in range(int(input())):
    n, x = rints()
    s, mem, ans = rstr(), [[0, 0] for _ in range(n)], 0
    zero, one = 0, 0

    for i in range(2):
        for j in range(n):
            mem[j][i] = zero - one
            if s[j] == '1':
                one += 1
            else:
                zero += 1
    for i, j in mem:
        if i == j:
            if i == x:
                ans = -1
                break
        else:
            t = (x - i) / (j - i)
            if int(t) == t and t > -1:
                ans += 1

    print(ans)