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 int n,x;
	public static int []a=new int[100000];
	public static String s;
	public static void Solve() {
		int ans=0;
		boolean infAns=false;
		int cntZeros=0;
		for(int i=0;i<n;i++) {
			if(s.charAt(i)=='0') {
				cntZeros++;
			}
		}
		int total=cntZeros-(n-cntZeros);
		int bal=0;
		for(int i=0;i<n;i++) {
			if(total==0) {
				if(bal==x) {
					infAns=true;
				}
			}else if(Math.abs(x-bal)%Math.abs(total)==0) {
				if((x-bal)/total>=0) {
					ans++;
				}
			}
			if(s.charAt(i)=='0') {
				bal++;
			}else {
				bal--;
			}
		}
		if(infAns)ans=-1;
		System.out.println(ans);
	}
	public static void main(String[]args) {
		Scanner sc=new Scanner(System.in);
		int T=sc.nextInt();
		while(T-->0) {
			n=sc.nextInt();
			x=sc.nextInt();
			s=sc.next();
			Solve();
		}
	}
}