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 JAVA):

Solution:
T=+readline()
function gcd(a,b){
    if (!b)
        return a;
    return gcd(b,a%b)
}
for (t=0;t<T;t++){
    s=readline().split(' ')
    n=+s[0]
    x=+s[1]
    s=readline()
    q=0
    for (i=0;i<=n;i++){
        if (i){
            if (s[i-1]=='1')
                --q
            else
                ++q
        }
    }
    bal=0
    ans=0
    for (i=0;i<n;i++){
        if (i){
            if (s[i-1]=='1')
                --bal
            else
                ++bal
        }
        if (q==0&&bal==x){
            ans=-1;
            break;
        }
        if ((x-bal)%q==0){
            if ((x-bal)/q>=0){
                ++ans
            }
        }
    }
    print(ans)
}