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.io.*;
import java.util.*;

public class CF1295B extends PrintWriter {
	CF1295B() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1295B o = new CF1295B(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int x = sc.nextInt();
			byte[] cc = sc.next().getBytes();
			int[] kk = new int[n + 1 + n];
			int d = 0, dmax = 0, dmin = 0;
			for (int i = 0; i < n; i++) {
				dmax = Math.max(dmax, d);
				dmin = Math.min(dmin, d);
				kk[n + d]++;
				if (cc[i] == '0')
					d++;
				else
					d--;
			}
			int ans = 0;
			if (d > 0) {
				if (x > dmax)
					x -= (x - dmax + d - 1) / d * d;
				while (x >= dmin) {
					ans += kk[n + x];
					x -= d;
				}
			} else if (d < 0) {
				d = -d;
				if (x < dmin)
					x += (dmin - x + d - 1) / d * d;
				while (x <= dmax) {
					ans += kk[n + x];
					x += d;
				}
			} else {
				if (dmin <= x && x <= dmax && kk[n + x] > 0)
					ans = -1;
			}
			println(ans);
		}
	}
}