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 Main {
	public static void main(String[] args) {
		InputReader in = new InputReader();
		StringBuilder out = new StringBuilder();

		int T = in.nextInt();
		while (T-- > 0) {
			int N = in.nextInt();
			int x = in.nextInt();
			char[] str = in.nextLine().toCharArray();
			int cnt = 0;
			Map<Integer, Integer> map = new HashMap<>();
			for (int i = 0; i < N; i++) {
				Integer val = map.get(cnt);
				map.put(cnt, val == null ? 1 : val + 1);
				if (str[i] == 48) {
					cnt++;
				} else {
					cnt--;
				}
			}
			int ans = 0;
			if (cnt == 0) {
				out.append((map.get(x) == null ? 0 : -1) + "\n");
				continue;
			}
			for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
				Integer key = entry.getKey();
				Integer val = entry.getValue();
				if ((x - key) % cnt == 0 && (x - key) / cnt >= 0) {
					ans += val;
				}
			}
			out.append(ans + "\n");
		}

		System.out.print(out);
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer st;

		public InputReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
		}

		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(nextLine());
			}
			return st.nextToken();
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}