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:
//make sure to make new file!
import java.io.*;
import java.util.*;

public class B1295{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      int t = Integer.parseInt(f.readLine());
      
      for(int q = 1; q <= t; q++){
      
         StringTokenizer st = new StringTokenizer(f.readLine());
      
         int n = Integer.parseInt(st.nextToken());
         int m = Integer.parseInt(st.nextToken());
         
         char[] array = f.readLine().toCharArray();
         
         int[] dif = new int[n];
         
         if(array[0] == '1') dif[0] = -1;
         else dif[0] = 1;
         
         HashSet<Integer> difs = new HashSet<Integer>();
         difs.add(dif[0]);
         for(int k = 1; k < n; k++){
            if(array[k] == '1') dif[k] = dif[k-1]-1;
            else dif[k] = dif[k-1]+1;
            difs.add(dif[k]);
         }
         
         if(dif[n-1] == 0){
            if(difs.contains(m)){
               out.println(-1);
            } else {
               out.println(0);
            }
         } else {
         
            int answer = 0;
            
            if(m == 0) answer++;
               
            if(dif[n-1] > 0){
               for(int k = 0; k < n; k++){
                  if(m >= dif[k] && (m-dif[k])%dif[n-1] == 0) answer++;
               }
            } else {
               for(int k = 0; k < n; k++){
                  if(m <= dif[k] && (dif[k]-m)%(dif[n-1]*-1) == 0) answer++;
               }
            }
            out.println(answer);
         }
            
      
      }
      
      
      
      
      out.close();
   }
   
      
}