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.math.*;
import java.io.*;

public class b {
	public b() {
		FS scan = new FS();
		PrintWriter out = new PrintWriter(System.out);
		int t = scan.nextInt();
		for(int tt = 0 ; tt < t ; tt++) {
			int n = scan.nextInt();
			long x = scan.nextLong();
			char[] s = scan.next().toCharArray();
			long[] prefix = new long[n];
			for(int i  = 0 ; i < n ; i++)
				prefix[i] = s[i] == '0' ? +1: -1;
			for(int i = 1 ; i < n ; i++)
				prefix[i] += prefix[i - 1];
			long total = prefix[n - 1];
			boolean infinite = false;
			long cnt = 0;
			for(int i = 0 ; i < n ; i++) {
				long dif = x - prefix[i];
				if(total == 0) {
					if(dif == 0) {
						infinite = true;
						break;
					}
					continue;
				}
				long occ = dif / total;
				if(occ < 0) continue;
				long test = (occ * total) + prefix[i];
				if(test == x) cnt++;
			}
			if(x == 0) cnt++; // empty prefix
			if(infinite) out.println(-1);
			else out.println(cnt);
		}
		out.close();
	}
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		public String next() {
			while(!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch(Exception e) { e.printStackTrace(); }
			}
			return st.nextToken();
		}
		public int nextInt() { return Integer.parseInt(next()); }
		public long nextLong() { return Long.parseLong(next()); }
		public double nextDouble() { return Double.parseDouble(next()); }
	}
	public static void main(String[] args) throws Exception { new b(); }
}
/*



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