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 InfinitePrefixes {

	// https://codeforces.com/contest/1295/problem/B
		
	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("InfinitePrefixes"));

		int t = Integer.parseInt(in.readLine());
		while (t --> 0) {
			StringTokenizer st = new StringTokenizer(in.readLine());
			int n = Integer.parseInt(st.nextToken());
			long x = Long.parseLong(st.nextToken());
			long ans=0;
			if (x == 0) ans++; // empty string
			
			char[] arr = in.readLine().toCharArray();
			long[] pref = new long[n];
			for (int i=0; i<n; i++) {
				if (i!=0 ) pref[i] = pref[i-1];
				if (arr[i] == '0') pref[i]++;
				else pref[i]--;
				
				if (pref[i] == x) ans++;
			}
			
			// first step, check 0+pref[i] (done above)
			// second step, check pref[n-1]*k + pref[i] = x
				// k = (x-pref[i]) / pref[n-1]
			
			if (ans>0 && pref[n-1] == 0) {
				System.out.println(-1);
				continue;    // infinite
			}
			else if (pref[n-1] == 0) {
				System.out.println(0);
				continue;
			}
			
			for (int i=0; i<n; i++) {
				if ((x-pref[i]>=0 == pref[n-1]>=0)) {	// same sign
					if (pref[n-1] == 0 && x - pref[i] == 0) {
						ans++;
					}
					else if (x != pref[i] && (x - pref[i])%pref[n-1] == 0) {
						ans++;
					}
				}
			}
			System.out.println(ans);
		}
	}
}