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.io.*;
import java.math.*;
 
   public class x1295B
   {
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         while(T-->0)
         {
            st = new StringTokenizer(infile.readLine());
            int N = Integer.parseInt(st.nextToken());
            int X = Integer.parseInt(st.nextToken());
            String input = infile.readLine();
            int[] arr = new int[N];
            for(int i=0; i < N; i++)
            {
               if(input.charAt(i) == '1')
                  arr[i]--;
               else
                  arr[i]++;
            }
            int res = 0;
            int total = 0;
            for(int i=0; i < N; i++)
            {
               total += arr[i];
               if(total == X)
                  res++;
            }
            if(total == 0)
            {
               if(res == 0)
                  sb.append(res);
               else
                  sb.append(-1);
               sb.append("\n");
            }
            else
            {
               //limited
               res = 0;
               int temp = 0;
               for(int i=0; i < N; i++)
               {
                  if(pos(total, temp, X))
                     res++;
                  temp += arr[i];
               }
               sb.append(res+"\n");
            }
         }
         System.out.print(sb);
      }
      public static boolean pos(int total, int small, int X)
      {
         if(X == small)
            return true;
         if(X-small > 0 && total < 0)
            return false;
         if(X-small < 0 && total > 0)
            return false;
         int dif = Math.abs(X-small);
         return dif%Math.abs(total) == 0;
      }
   }