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 collections import Counter


def solve(n, x, s):
    assert len(s) == n
    # Track the balance seen for one cycle
    balance = 0
    seen = Counter()
    for c in s:
        balance += 1 if c == "0" else -1
        seen[balance] += 1

    if balance == 0:
        # Balance equal 0 at end of one cycle, same values will repeat forever
        return -1 if x in seen else 0
    else:
        # On the next cycle all the values in seen will appear again except += balance.
        # This is the same as looking for the value of (x - balance) in seen instead.
        # In particular in the (k+1)th cycle we can look for (x - k * balance) in seen.
        # Note: There are some ugly edge cases to handle the cycling of seen which
        # is exclusive of start point (the very first balance == 0).
        minSeen = min(0, min(seen.keys()))
        maxSeen = max(0, max(seen.keys()))
        ans = 1 if x == 0 else 0

        # If x is big, skip k loops ahead to get it into the range of seen
        if balance < 0:
            k = (x - minSeen) // balance
        if balance > 0:
            k = (x - maxSeen) // balance
        x -= max(0, k) * balance
        # Iterate through (x - k * balance) until out of range
        while True:
            ans += seen[x]
            x -= balance
            if balance < 0 and x > maxSeen:
                break
            if balance > 0 and x < minSeen:
                break
        return ans


T = int(input())

for t in range(T):
    n, x = map(int, input().split())
    s = input().rstrip()
    print(solve(n, x, s))