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)
	{
		Scanner s=new Scanner(System.in);
		
		int t=s.nextInt();
		
		for(int i=0;i<t;i++)
		{
			int n=s.nextInt();
			long x=s.nextInt();
			
			String str=s.next();
			
			long[] dp=new long[n];
			
			if(str.charAt(0)=='0')
				dp[0]=1;
			else
				dp[0]=-1;
			
			for(int j=1;j<n;j++)
			{
				if(str.charAt(j)=='0')
				{
					dp[j]=dp[j-1]+1;
				}
				else
				{
					dp[j]=dp[j-1]-1;
				}
			}
			
			long count=0;
			
			int p=0;
			
			for(int j=0;j<n;j++)
			{
				long val=x;
				val=val-dp[j];
				
				if(dp[n-1]==0)
				{
					if(val==0)
					{
						p=1;
						break;
					}
				}
				else
				{
					if(val%dp[n-1]==0&&(val/dp[n-1])>=0)
					{
						count++;
					}
				}
			}
			
			if(x==0)
				count++;
			
			if(p==0)
				System.out.println(count);
			else
				System.out.println(-1);
			
		}
		
	}
	
}