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 Edu_81B {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		StringTokenizer st;
		
		int q = sc.nextInt();
		while(q-->0) {
			int n = sc.nextInt(), x = sc.nextInt();
			String s = sc.next();
			
			int[] sum = new int[n];
			int f = 0;
			boolean flag = (x == 0 ? true: false);
			for(int i = 0; i < n; i++) {
				sum[i] = f + (s.charAt(i) == '0' ? 1 : -1);
				f = sum[i];
				if(f == x)
					flag = true;
			}
			
			if(f == 0) {
				if(flag) {
					System.out.println(-1);
				} else {
					System.out.println(0);
				}
			} else {
				int count = (x == 0 ? 1 : 0);
				for(int i = 0; i < n; i++) {
					int z = x - sum[i];
					if(z == 0) {
						count++;
					} else {
						if((z < 0 && f < 0) || (z > 0 && f > 0)) {
							if(z % f == 0) {
								count++;
							}
						}
					}	
				}
				System.out.println(count);
			}
		}
	}
}