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 math
import sys

def listinput(): return list(map(int,input().split(' ')))

def main():
    answer = ""
    for test in range(int(input())):
        n,x = listinput()
        s = input()
        mset = dict()
        balance = 0
        isin = False
        mset[0] = 1
        
        for letter in s:
            if letter == "0":
                balance += 1
            else:
                balance -= 1
            if mset.get(balance,None) == None:
                mset[balance] = 1
            else:
                mset[balance] += 1
            if balance == x:
                isin = True


        mset[balance] -= 1
        am_of_x = 0
        if balance == 0:
            if isin:
                answer += "-1 \n" 
            else:
                answer += "0 \n"
        else:
            #for key: if x = key + balance*i where i is natural
            for key in mset.keys():
                if (x-key)%balance == 0 and (x-key)//balance >= 0:
                    am_of_x += mset[key]
                
            answer += str(am_of_x) + "\n"
        
        pass
    print(answer)

main()