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 sc = new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			int x=sc.nextInt();
			String s=sc.next();
			int a[]=new int[n+10];
			a[0]=0;
			int ans=0;
			for(int i=1;i<=n;i++) {
				if(s.charAt(i-1)=='1') a[i]=a[i-1]-1;
				else a[i]=a[i-1]+1;
			}
			for(int i=0;i<=n;i++) {
				if(a[i]==x) {
					ans++;
					if(a[n]==0) {ans=-1;break;}
				}
				else {
					if(a[n]!=0&&i!=0) {
						if(a[i]<x&&a[n]>0&&(x-a[i])%a[n]==0) ans++;
						if(a[i]>x&&a[n]<0&&(a[i]-x)%a[n]==0) ans++;
					}
				}
			}
			System.out.println(ans);
		}
		sc.close();
	}
}
//1 0 1 2 1 2
// 10