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 Main{
	
	 public static void main (String args[]){
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int t = sc.nextInt();
        while(t-->0) {
        	int n = sc.nextInt();
        	int x = sc.nextInt();
        	char arr[] = sc.next().toCharArray();
        	int arr1[] = new int[2*n+1], max=Integer.MIN_VALUE, min=Integer.MAX_VALUE, diff=0, count=0;
        	
        	for(int i=0; i<n; i++) {
        		if(arr[i]=='0')
        			diff++;
        		else
        			diff--;
        		if(diff > max)
        			max = diff;
        		if(diff<min)
        			min=diff;
        		arr1[n+diff]++;
        	}
        	
        	if(x==0 && diff!=0 )
        		count++;
        	if(diff>0) {
        		if(x>max)
        			x -= (x-max+diff-1)/diff*diff;
        		while(x>=min) {
        			count+=arr1[n+x];
        			x-=diff;
        		}
        	}
        	else if(diff<0) {
        		diff=-diff;
        		if(x<min)
        			x += (min-x+diff-1)/diff*diff;
        		while(x<=max) {
        			count+=arr1[n+x];
        			x+=diff;
        		}
        	}else {
        		if(x>=min && x<=max && arr1[n+x]>0)
        			count=-1;
        	}
        	
        	out.println(count);
        	out.flush();
        } 
    }
}