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 Solution
{
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
              int t = sc.nextInt();
              while(t-- > 0)
              {
                     int n = sc.nextInt();
                     int x = sc.nextInt();
                     sc.nextLine();
                     String s = sc.nextLine();
                     
                     int cnt[] = new int[n];
                     int sum = 0;
                     for(int i = 0 ; i < n ; i++)
                     {
                            if(s.charAt(i) == '0')
                            sum++;
                            
                            else
                            sum--;
                            
                            cnt[i] = sum;
                     }
                     
                     if(sum == 0)
                     {
                            boolean found = false;
                            for(int i = 0 ; i < n ; i++)
                            {
                                   if(x == cnt[i])
                                   found = true;
                            }
                            if(found)
                            System.out.println(-1);
                            
                            else
                            System.out.println(0);
                     }
                     
                     else
                     {
                         int tot = 0;
                         for(int i = 0 ; i < n ; i++)
                         {
                               int a = x-cnt[i];
                               int b = sum;
                               
                               if(Math.abs(a)%Math.abs(b) == 0)
                               {
                                      if((a >= 0 && b >= 0) || (a <= 0 && b <= 0))
                                      tot++;
                               }
                                
                         }
                         if(x == 0)
                         tot++;
                         
                         System.out.println(tot);
                     }
              }
       }
}