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.math.*;
import java.util.*;
import java.io.*;
  
public class B { 
  
    static BufferedReader in;
    static String file = "../in";
    static int test = 10; // 0 for local testing, 1 for std input

    public static void main(String[] args) throws Exception
    {
        int _k = Integer.valueOf("1");
        if(test > 0) in = new BufferedReader(new InputStreamReader(System.in));
        else in = new BufferedReader(new FileReader(file));
        if(test < 0) {String[] str = in.readLine().split(" ");}

        
        
        /****************************************************/
        /****************************************************/
        /****************************************************/
        /****************************************************/

        int t = Integer.valueOf(in.readLine());
        while(t-- > 0)
        {
            
            String[] sp = in.readLine().split(" ");
            
            int n = Integer.valueOf(sp[0]),
                x = Integer.valueOf(sp[1]);

            String str = in.readLine();
            
            int sum = 0;
            int[] count = new int[n + 1];
            
            for(int i = 0; i < n; i++)
            {
                if(str.charAt(i) == '0') count[i + 1] = count[i] + 1;
                else count[i + 1] = count[i] - 1;
            }
            // if(test == 0) 
                // System.out.println(Arrays.toString(count));

            int last = count[n];    // will increase by this amount

            int re = 0;
            if(x == 0) re++;
            boolean inf = false;

            for(int i = 0; i < n; i++)
            {
                int diff = x - count[i + 1];
                if(last == 0)
                {
                    if(diff == 0)
                    {
                        System.out.println("-1");
                        inf = true;
                        break;
                    }
                }
                else if(diff % last == 0)
                {
                    
                    int op = diff / last;
                    // if(op == 0) // infinite
                    // {
                    //     System.out.println("-1");
                    //     break;
                    // }
                    if(op >= 0) re++;
                }
            }
            if(!inf) System.out.println(re);

        }
        
    }
}