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.lang.*;

public class InfinitePrefixes {

	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
		int t=scn.nextInt();
		while(t-- !=0) {
			
			int n=scn.nextInt();
			int x=scn.nextInt();
			scn.nextLine();
			String s=scn.nextLine();
			
			int[]arr=new int[n];
			int count=0;
			
			for(int i=0;i<n;i++) {
				char ch=s.charAt(i);
				if(ch=='1')
					count--;
				else
					count++;
				arr[i]=count;
			}
			
			if(count==0) {
				boolean found=false;
				for(int i=0;i<n;i++)
					if(arr[i]==x)
					{
						System.out.println("-1");
						found=true;
						break;
					}
				if(!found)
					System.out.println("0");
				continue;
			}else {
				int ans=0;
				if(x==0)
					ans++;
				for(int i=0;i<n;i++) {
					
					if((x-arr[i])%arr[n-1]==0 && ((x-arr[i])/arr[n-1])>=0 ) 
						ans++;
				}
				System.out.println(ans);
			}
		}
	}

}