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 B {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
			Scanner in =new Scanner(System.in);
			int q =in.nextInt();
			for(int p=0;p<q;p++) {
				int len =in.nextInt();
				int b=in.nextInt();
				String x= in.next();
			char c[]=x.toCharArray();
			int count =0;int max =0,min=0;
			for(int i=0;i<c.length;i++) {
				if(c[i]=='0') {
					count++;
					max=Math.max(max, count);
				}else {
					count--;min=Math.min(min, count);
				}
			}
			if(count==0) {
				if(b<=max&&b>=min) {
					System.out.println("-1");
				}else {
					System.out.println("0");
				}
			}else {
				
				int count2=0;int cnt=0;
				if(b==0) {
					cnt++;
				}
				for(int i=0;i<c.length;i++) {
					if(c[i]=='0') {
						count2++;
						
						if((b-count2)%count==0&&((((b-count2)>=0)&&(count>=0))||(((b-count2)<=0)&&(count<=0)))) {
							cnt++;
						}
					}else {
						count2--;
						if((b-count2)%count==0&&((((b-count2)>=0)&&(count>=0))||(((b-count2)<=0)&&(count<=0)))) {
							cnt++;
						}
					}
				}
				System.out.println(cnt);
			}
			}
	}

}