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

public class GFG {
    public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(br.readLine().trim());
		while(t--!=0){
    		String s=br.readLine().trim();
    		String ss[]=s.split("\\s+");
    		int n=Integer.parseInt(ss[0]);
    		int k=Integer.parseInt(ss[1]);
    		s=br.readLine().trim();
    		int d[]=new int[n+1];
    		for(int i=1;i<=n;i++){
    		    char c=s.charAt(i-1);
    		    if(c=='0')
        		    d[i]=d[i-1]+1;
        		else
        		    d[i]=d[i-1]-1;
    		}
    		int cnt=0;
    		if(d[n]==0){
    		    for(int i=1;i<=n;i++){
    		        if(d[i]==k){
    		            cnt=-1;
    		            break;
    		        }
    		    }
    		    System.out.println(cnt);
    		    continue;
    		}
    		
    		if(k==0)
    		    cnt=1;
    		for(int i=1;i<=n;i++){
    		    if(((k-d[i])%d[n]==0&&(k-d[i])/d[n]>=0))
    		        cnt++;
    		}
    		System.out.println(cnt);
		}
	}
}