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 collections import defaultdict

def solve(s, x):
    remains = defaultdict(int)
    remains[0] = 1
    s = map(int, s)
    b = 0
    for num in s:
        if num == 0:
            b += 1
        else:
            b -= 1
        remains[b] += 1
    remains[b] -= 1

    res = 0
    for r, c in remains.items():
        y = x - r
        if b == 0:
            if y == 0:
                return -1
        elif y % b == 0 and y / b >= 0:
            res += c
    return res

assert solve('010010', 10) == 3
assert solve('10101', 3) == 0
assert solve('0', 0) == 1
assert solve('01', 0) == -1

T = int(raw_input())

for case_ in xrange(T):
    n, x = map(int, raw_input().split())
    s = raw_input()
    print solve(s, x)