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.io.*;

public class contest {
	
	public static void main(String[]args)throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		
		int t = Integer.parseInt(br.readLine());
		while(t-- > 0)
		{
			StringTokenizer st1 = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st1.nextToken());
			long x = Long.parseLong(st1.nextToken());
			String ss = br.readLine();
			long prefix[] = new long[n];
			int cntz = 0;
			int cnto = 0;
			for(int i = 0 ; i < n ; i++)
			{
				if(ss.charAt(i) == '0')
					cntz++;
				else
				{
					cnto++;
				}
				prefix[i] = cntz-cnto;
			}
			int res = 0;
			boolean flag = true;
			for(int i = 0 ; i < n ; i++)
			{
				if(x-prefix[i] == 0 )
				{
					if(prefix[n-1] == 0)
					{
						flag = false;
						break;
						
					}else {
						res++;
					}
				}else {
					if(prefix[n-1] != 0 && (x-prefix[i]) % prefix[n-1] == 0 && (Long.signum(x-prefix[i]) == Long.signum(prefix[n-1])))
						res++;
				}
			}
			
			if(flag )
			{
				if(x == 0)
					res++;
				out.println(res);
			}
			else
				out.println(-1);
			
		}
		
		out.flush();
		out.close();
	}

}