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:
/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;
import java.lang.*;

public class GFG {
	public static void main (String[] args) {
	    FastReader fr = new FastReader();
	    int t = fr.nextInt();
	    while(t-->0){
	        int n = fr.nextInt(),x = fr.nextInt();
	        String s = fr.nextLine();
	        long cnt=0;
	        for(int i=0;i<n;i++){
	            if(s.charAt(i)=='0') cnt++;
	            else cnt--;
	        }
	        long ans = 0;
	        StringBuilder st = new StringBuilder();
	        
	        if(cnt==0){
	            for(int i=0;i<n;i++){
	                if(cnt==x){
	                    ans = -1;break;
	                }
	                if(s.charAt(i)=='0') cnt++;
	                else cnt--;
	            }
	        }
	        else{
	            long tt = cnt;
	            
	            cnt =0;
	            for(int i=0;i<n;i++){
	                long diff = x-cnt;
	                if(diff%tt==0&&diff/tt>=0) ans++;
	                if(s.charAt(i)=='0') cnt++;
	                else cnt--;
	            }
	        }
	        System.out.println(ans);
	    }
	}
}
class FastReader {
 
    BufferedReader bf;
    StringTokenizer st;
 
    public FastReader() {
        bf = new BufferedReader(new InputStreamReader(System.in));
    }
 
    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(bf.readLine());
            } catch (Exception e) {
                System.out.println(e);
            }
        }
        return st.nextToken();
    }
 
    int nextInt() {
        return Integer.parseInt(next());
    }
 
    long nextLong() {
        return Long.parseLong(next());
    }
 
    double nextDouble() {
        return Double.parseDouble(next());
    }
 
    String nextLine() {
        String str = "";
        try {
            str = bf.readLine();
        } catch (Exception e) {
            System.out.println(e);
        }
        return str;
    }
}