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.io.*;
import java.util.*;
 
/**
 * Only By Abhi_Valani
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Unstoppable solver = new Unstoppable();
		int t=in.nextInt();
	    while(t-->0)
        solver.solve(in, out);
        out.close();
    }
 
    static class Unstoppable {
        public void solve(InputReader in, PrintWriter out)
        {
          int n=in.nextInt();
           int x=in.nextInt();
          char ch[]=in.next().toCharArray();
          int c1=0,c0=0;
          int diff[]=new int[n];
          for(int i=0;i<n;++i)
          {
              if(ch[i]=='1') c1++;
              else c0++;
              diff[i]=c0-c1;
          }
          if(diff[n-1]==0)
          {
              int flag=0;
          for(int i=0;i<n;++i)
          {
              if(x==diff[i]) flag=1;
          }
          if(flag==1) out.println("-1");
          else if(x==0) out.println("1");
          else{
              out.println("0");
          }
          }
          
          else{ int count=0;
              for(int i=0;i<n;++i)
          {
              double temp=(x-diff[i])/(double)diff[n-1];
             // out.println(temp);
              if(temp>=0&&(temp-(long)temp)==0) count++;
          } if(x==0) count++;
          out.println(count);
          }
        }
 
    }
 
 
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
 
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
 
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
    }
}