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:
/*
javac a.java && java a
*/

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

public class a {
	public static void main(String[] args) { new a(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);

	int t;
	int n, x;
	char[] s;
	int[] pb;

	a() {
		t = in.nextInt();
		while (t-- > 0) {
			n = in.nextInt();
			x = in.nextInt();
			s = in.next().toCharArray();
			pb = new int[n + 1];
			for (int i = 1; i <= n; i++) {
				pb[i] = pb[i - 1];
				if (s[i - 1] == '0') pb[i]++;
				else pb[i]--;
			}
			int cnt = 0;
			boolean inf = false;
			if (pb[0] == x) cnt++;
			for (int i = 1; i <= n; i++) {
				if (pb[n] == 0 && pb[i] == x) inf = true;
				else if (pb[i] == x) cnt++;
				else {
					if (pb[n] == 0) continue;
					int del = x - pb[i];
					if (del < 0 && pb[n] < 0)
						if (abs(del) % abs(pb[n]) == 0) cnt++;
					if (del > 0 && pb[n] > 0)
						if (del % pb[n] == 0) cnt++;
				}
			}
			if (inf) out.println(-1);
			else out.println(cnt);
		}
		out.close();
	}

	
	
	int abs(int x) { if (x < 0) return -x; return x; }
	long abs(long x) { if (x < 0) return -x; return x; }
	int min(int a, int b) { if (a < b) return a; return b; }	
	int max(int a, int b) { if (a > b) return a; return b; }	
	long min(long a, long b) { if (a < b) return a; return b; }	
	long max(long a, long b) { if (a > b) return a; return b; }	
	int gcd(int a, int b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
	long gcd(long a, long b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
	long lcm(int a, int b) { return (((long) a) * b) / gcd(a, b); }
	long lcm(long a, long b) { return (a * b) / gcd(a, b); }

	void sort(int[] arr) {
		int sz = arr.length, j;
		Random r = new Random();
		for (int i = 0; i < sz; i++) {
			j = r.nextInt(i + 1);
			arr[i] = arr[j]^(arr[i]^(arr[j] = arr[i]));
		} Arrays.sort(arr);
	}

	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}