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:
/*package whatever //do not write package name here */

import java.util.*;

public class GFG {
	public static void main (String[] args) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while(t-->0)
		{
		    int n = in.nextInt();
		    int x = in.nextInt();
		    String s = in.next();
		    System.out.println(getAns(n,x,s));
		}
	}
	static int getAns(int n, int x, String s)
	{
	    int bal[] = new int[n+1];
	    int cnt0 = 0;
	    for(int i = 0;i<n;i++)
	    {
	        if(s.charAt(i)-'0'==0)
	        cnt0++;
	        bal[i+1] = 2*cnt0 - (i+1); 
	    }
	   // System.out.println(Arrays.toString(bal));
	    if(bal[n]==0)
	    {
	        for(int i = 0;i<=n;i++)
	        {
	            if(bal[i]==x)
	            return -1;
	        }
	        return 0;
	    }
	    else
	    {
	        int cnt =0 ;
	        for(int i =1;i<=n;i++)
	        {
	            if((x-bal[i])%bal[n]==0&&((x-bal[i])/bal[n])>=0)
	            {
	                cnt++;
	               // System.out.println(i);
	            }
	        }
	        if(x==0)
	        cnt++;
	        return cnt;
	    }
	
	}
}