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.io.*;

public class b {
	
	public static int n;
	public static int target;
	public static int savetarget;
	
	public static void main(String[] args) throws Exception {
	
		BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
		int nC = Integer.parseInt(stdin.readLine().trim());
		
		for (int loop=0; loop<nC; loop++) {
			StringTokenizer tok = new StringTokenizer(stdin.readLine());
			
			n = Integer.parseInt(tok.nextToken());
			target = Integer.parseInt(tok.nextToken());
			savetarget = target;
			String s = stdin.readLine().trim();
			
			// Get all prefixes within the string.
			int[] cnt = new int[n+1];
			for (int i=0; i<n; i++) {
				int add = s.charAt(i) == '0' ? 1 : -1;
				cnt[i+1] = cnt[i] + add;
			}
			
			// Get a frequency array here.
			int[] freq = new int[2*n+1];
			for (int i=1; i<cnt.length; i++)
				freq[n+cnt[i]]++;
			
			System.out.println(solve(cnt, freq));
			
		}
	
	}
	
	public static int solve(int[] cnt, int[] freq) {
		
		int minF = 0, maxF = 0;
		for (int i=0; i<cnt.length; i++) {
			minF = Math.min(minF, cnt[i]);
			maxF = Math.max(maxF, cnt[i]);
		}
		int d = cnt[n];
		
		// Means we always balance out at the end.
		if (d == 0) {
			if (target >= minF && target <= maxF) return -1;
			else return 0;
		}
		
		// Finite case.
		
		if (target > maxF && d < 0) return 0;
		if (target < minF && d > 0) return 0;
		
		if (d > 0) {
			
			if (target > maxF) {
				int subOut = (target-maxF+d-1)/d;
				target -= (subOut*d);
			}
			
			int res = 0;
			while (target >= minF) {
				res += freq[target+n];
				target -= d;
			}
			
			if (savetarget == 0) res++;
			
			return res;
			
		}
		else {
			
			if (target < minF) {
				int subOut = (minF-target-d-1)/(-d);
				target -= (subOut*d);
			}
			
			int res = 0;
			while (target <=maxF) {
				res += freq[target+n];
				target -= d;
			}
			
			if (savetarget == 0) res++;
			
			return res;
			
		}
	}
}