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:
from itertools import accumulate
from collections import defaultdict

T = int(input())
for _ in range(T):
    n, x = map(int, input().split())
    s = [1 if c == '0' else -1 for c in input()]
    accum = list(accumulate(s))
    count = defaultdict(int)
    for i in accum:
        count[i] += 1
    last = accum[-1]
    ans = 0
    if x == 0:
        ans = 1
    if last == 0:
        if count[x] > 0:
            ans = -1
    else:
        ma, mi = max(accum), min(accum)
        if last > 0:
            x -= max(0, (x-ma+last-1)//last * last)
        else:
            x += max(0, (mi-x+last-1)//last * last)
        while (last > 0 and x >= mi) or (last < 0 and x <= ma):
            ans += count[x]
            x -= last
    print(ans)