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 B {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int T = sc.nextInt();
		PrintWriter pw = new PrintWriter(System.out);
		for(int _t = 0; _t < T; _t++) {
			int n = sc.nextInt();
			long x = sc.nextInt();
			char[] s = sc.next().toCharArray();
			int[] bal = new int[n+1];
			for(int i = 0; i < n; i++) {
				bal[i+1] = bal[i] + (s[i] == '0'? 1:-1);
			}
			long d = bal[n];
			if(d == 0) {
				long res = 0;
				for(int i = 0; i < n; i++) {
					if(bal[i] == x) res++;
				}
				if(res > 0) {
					System.out.println("-1");
				}
				else {
					System.out.println("0");
				}
			}
			else {
				long res = 0;
				for(int i = 0; i < n; i++) {
					if(compare(bal[i], x, d)) res++;
				}
				System.out.println(res);
			}
		}
		pw.flush();
	}
	static boolean compare(long bal, long x, long d) {
		if(d > 0 && x < bal) return false;
		if(d < 0 && x > bal) return false;
		if(d < 0) {
			bal = -bal;
			x = -x;
			d = -d;
		}
		bal = (bal % d + d) % d;
		x = (x % d + d) % d;
		return bal == x;
	}
	
	static class FastScanner {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
		public FastScanner() {
			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
			tokenizer = null;
		}
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch(IOException e) {
				throw new RuntimeException(e);
			}
		}
	}


}