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 InfinitePrefixes {
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int test = sc.nextInt();
		
		while(test-- > 0) {
			int n = sc.nextInt();
			int x = sc.nextInt();
			String s = sc.next();
			
			long cnt = 0;
			
			for(int i = 0; i < n; i++) {
				if(s.charAt(i) == '0') {
					cnt++;
				}
				else {
					cnt--;
				}
			}
			
			long ans = 0;
			
			if(cnt == 0) {
				for(int i = 0; i < n; i++) {
					if(cnt == x) {
						ans = -1;
						break;
					}
					
					if(s.charAt(i) == '0') {
						cnt++;
					}
					else {
						cnt--;
					}
				}
			}
			
			else {
				long tt = cnt;
				cnt = 0;
				
				for(int i = 0; i < n; i++) {
					long diff = x - cnt;
					if(diff % tt == 0 && diff / tt >=0) {
						ans++;
					}
					
					if(s.charAt(i) == '0') {
						cnt++;
					}
					else {
						cnt--;
					}
				}
			}
			System.out.println(ans);
		}
	}
}