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.*;

public class b {
  static boolean debug = false;
  public static void main(String[] args) {
    FS in = new FS(System.in); 

    int numT = in.nextInt();
    for(int T=0; T<numT; T++) {
      int n = in.nextInt();
      int x = in.nextInt();

      char[] word = in.next().toCharArray();
      int[] val = new int[n];
      int[] freq = new int[2];
      int[] offset = new int[n];
      // int[] freqOffset = new int[n];
      boolean possibleInFirst = false;
      for(int i=0; i<n; i++) {
        val[i] = (int)(word[i] - '0');
        freq[val[i]]++;
        offset[i] = (i==0) ? 0 : offset[i-1];
        offset[i] += (val[i] == 0) ? 1 : -1;
        // freqOffset[offset[i]]++;
        if(offset[i] == x) possibleInFirst = true;
        if(debug) System.out.printf("%d ", offset[i]);
      }
      if(debug) System.out.println();
      
      // offset of whole word:
      int wordOffset = freq[0] - freq[1];
      if(debug) System.out.printf("wordoffset=%d\n", wordOffset);
      int ans = 0;
      int target = x;
      if(freq[0] == freq[1]) {
        // always possible if possible in first
        System.out.println((possibleInFirst) ? "-1" : "0");
      } else {
        for(int i=0; i<n; i++) {
          int distToRight = target - offset[i];
          if((distToRight < 0 && wordOffset > 0) || (distToRight > 0 && wordOffset < 0)) {
            continue;
          }
          if(distToRight%wordOffset == 0) {
            ans++;
          }
        }
        if(x == 0) ans++; // for empty prefix
        System.out.println(ans);
      }


    }
  }

  static class FS {
    BufferedReader in;
    StringTokenizer token;
    public FS(InputStream stream) {
      in = new BufferedReader(new InputStreamReader(stream));
    }
    public String next() {
      if(token == null || !token.hasMoreElements()) {
        try {
          token = new StringTokenizer(in.readLine());
        } catch (Exception e) {} 
        return next();
      }
      return token.nextToken();
    }
    public int nextInt() {
      return Integer.parseInt(next());
    }
  }
}

/* 


4
6 10
010010
5 3
10101
1 0
0
2 0
01

3
0
1
-1


1
1 0
0

1


1
5 3
10101

0


010010010010010010010010010010010010010010
    1  2  3  4  5  6  7  8 =+= =


Infinite if the # 0s and #1s are = for the string and it can be done within string.
If not possible in string its 0 if no offset standard.

*/