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.lang.*;
import java.io.*;

public class B {
	public static void main (String[] args) throws java.lang.Exception {
		new Solution();
	}
}
class Solution {
	Scanner scanner;
	public Solution() {
		scanner = new Scanner(System.in);
		for (;scanner.hasNext();) {
			int T = Integer.parseInt(scanner.nextLine());
			for (int i=0; i<T; i++) {
				run_case();
			}
		}
	}

	private void run_case() {
		String tc1 = scanner.nextLine();
		int n = Integer.parseInt(tc1.split("\\s+")[0]);
		int x = Integer.parseInt(tc1.split("\\s+")[1]);
		String s = scanner.nextLine();

		int res = 0;

		Map<Integer, Integer> map_cnt = new HashMap<>();
		int cur_score = 0;

		// map_cnt.put(cur_score, 1);
		if (x == 0) res ++;

		for (char c: s.toCharArray()) {
			if (c == '0') cur_score ++;
			else cur_score --;

			map_cnt.putIfAbsent(cur_score, 0);
			map_cnt.put(cur_score, map_cnt.get(cur_score) + 1);

			if (cur_score == x) res ++;
		}

		if (cur_score == 0 && map_cnt.containsKey(x)) {
			System.out.println("-1");
		} else {
			int r = cur_score;
			if (r != 0) {
				for (int key : map_cnt.keySet()) {
					if ( ((x - key) % r == 0) && (x - key) / r > 0) {
						res += map_cnt.get(key);
					}
				}
			}
			System.out.println(res);
		}
	}
}

// a = ()
// r = cur_score
 // tar - a % r = 0
// 1 0 1 2 1 2
// 3 2 3 4 3 4
//

// 0 -> 2
// 1 -> 3
// 2 -> 2