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:
import java.io.*;
import java.util.*;

public class B
{
    void solve(FastIO io)
    {
        int n = io.nextInt();
        int k = io.nextInt();

        String s = io.next();
        int[] a = new int[n];

        a[0] = s.charAt(0) == '0' ? 1 : -1;

        for (int i = 1; i < n; i++)
            a[i] = a[i-1] + (s.charAt(i) == '0' ? 1 : -1);

        int total = a[n-1];
        int cnt = 0;

        if (total == 0)
        {
            for (int i = 0; i < n; i++)
                if (a[i] == k)
                {
                    io.println(-1);
                    return;
                }

            if (k != 0)
                io.println(0);
            else
                io.println(-1);
            return;
        }
        else if (total == 1 || total == -1)
        {            
            for (int i = 0; i < n; i++)
                if ((total == 1 && a[i] <= k) || (total == -1 && a[i] >= k))
                    cnt++;

            if (k == 0)
                cnt++;

            io.println(cnt);
            return;
        }
        
        if (k == 0)
            cnt++;

        for (int i = 0; i < n; i++)
            if ((k - a[i]) % total == 0 && (k - a[i]) / total >= 0)
                cnt++;

        io.println(cnt);
    }

    public static void main(String[] args) 
    {
        FastIO io = new FastIO();

        int t = io.nextInt();

        for (int i = 0; i < t; i++)
            new B().solve(io);

        io.close();
    }

    static class FastIO extends PrintWriter
    {
        StringTokenizer st = new StringTokenizer("");
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));

        FastIO()
        {
            super(System.out);
        }

        String next()
        {
            while (!st.hasMoreTokens())
            {
                try {
                    st = new StringTokenizer(r.readLine());
                } catch (Exception e) { }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }
    }
}