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:
import sys

t = int(sys.stdin.readline())

def neg(s):
    ans = ['0']*len(s)
    for i,c in enumerate(s):
        if c=='0':
            ans[i] = '1'
    return ans

for _ in range(t):
    n,x = [int(i) for i in sys.stdin.readline().split()]
    s = sys.stdin.readline().strip()
    at = 0
    if x<0:
        x=-x
        s = neg(s)
    gain = len(list(filter(lambda c: c=='0', s))) - len(list(filter(lambda c: c=='1',s)))
    balance = 0
    ans = 0
    for c in s:
        if gain==0:
            if balance==x:
                ans=-1
                break
        else:
            if not (balance>x and gain>0 or balance<x and gain<0):
                if (balance-x)%gain == 0:
                    ans += 1
        if c=='0':
            balance+=1
        else:
            balance-=1
    print(ans)