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.util.*;
public class GFG {
   public static void main (String[] args) {
		Scanner sc=new Scanner(System.in);
	    int t1=sc.nextInt();
	    while (t1-->0)
	    {
	        int n=sc.nextInt();
	        int x=sc.nextInt();
	        String s=sc.next();
	        int pre[]=new int[n];
	        if(s.charAt(0)=='0')
	        pre[0]=1;
	        if(s.charAt(0)=='1')
	        pre[0]=-1;
	        for(int i=1;i<n;i++)
	        pre[i]=pre[i-1]+((s.charAt(i)=='0')?1:-1);
	        if(pre[n-1]==0)
	        {
	            int f=0;
	            for(int i=0;i<n;i++)
	            {
	                if(pre[i]==x){
	                System.out.println(-1);f=1;break;}
	            }
	            if(f==0)
	            System.out.println(0);
	            continue;
	        }
	        int ans=0;
	        for(int i:pre)
	        {
	            if(pre[n-1]!=0&&(x-i)%pre[n-1]==0&&(x-i)/pre[n-1]>=0)
	            ans++;
	        }
	        if(x==0)
	        ans++;
	        System.out.println(ans);
	    }
   }
}