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.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn = new Scanner(System.in);
		int t = scn.nextInt();
	  while(t-->0){
		  int n =scn.nextInt();
		  int x =scn.nextInt();
		  String s = scn.next();
		  System.out.println(res(n,s,x));
	  }

	}
	public static int res(int n , String s,int x){
		int count=0;
		for(int i=0;i<s.length();i++){
			char ch = s.charAt(i);
			if(ch=='0'){
				count++;
			}else{
				count--;
			}
		}
		int b=0;
		int ans=0;
	for(int i=0;i<s.length();i++){
	     if(count==0){
	    	 if(x==b){
	    		 return -1;
	    	 }
	     }
	     else if((int)Math.abs(x-b)%(int)(Math.abs(count))==0){
	    	 if(((x-b)/count)>=0){
	    		 ans++;
	    	 }
	     }
	     if(s.charAt(i)=='0'){
	    	 b++;
	     }else{
	    	 b--;
	     }
	}
		
		return ans;
	}

}