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.*;
import java.util.Arrays;
public class solve
{
    public static void main(String args[])
    {
        Scanner in= new Scanner(System.in);
        int T=in.nextInt();
       while(T!=0)
       {
           T--;
          
           int n=in.nextInt();
           int x=in.nextInt();
           String str=in.next();
           int a[]=new int[n+1];
           int b[]=new int[n+1];
           int s=0;
           
           for(int i=1;i<=n;i++)
           {
               if(str.charAt(i-1)=='0')
               s++;
               else
               s--;
               a[i]=s;
           }
           int max=a[0],min=a[0];
           for(int i=1;i<=n;i++)
           {
               b[i]=x-a[i];
               
           }
           int c=0;
           if(x==0)
           c=1;
           if(a[n]==0)
           {
           for(int i=1;i<n+1;i++)
           {
               if(a[i]==x)
               {
               c=-1;
               break;
               }
               
           }
           }
           else if(a[n]>0)
           {
               for(int i=1;i<=n;i++)
               {
                   if((b[i]%a[n]==0)&&(b[i]>=0))
                   c++;
               }
           }
           else
           {
           for(int i=1;i<=n;i++)
               {
                   if((b[i]%a[n]==0)&&(b[i]<=0))
                   c++;
               }
           } 
           System.out.println(c);
           
           
           
           
           
       }
    }
    
}