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 Solution {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
	public static void main(String args[]) throws IOException {
		int t = Integer.parseInt(br.readLine());
		while(t-- > 0) {
	        String inp[] = br.readLine().split(" ");
	        int n = Integer.parseInt(inp[0]);
	        int x = Integer.parseInt(inp[1]);
	        String str = br.readLine();
	        int cnt0 = 0, cnt1 = 0;
	        int diff[] = new int[n];
	        for(int i = 0; i < n; i++) {
	            char ch = str.charAt(i);
	            if(ch == '0') {
	                cnt0++;
	            } else {
	                cnt1++;
	            }
	            diff[i] = cnt0 - cnt1;
	        }
	        int res = 0;
	        int end = diff[n - 1];
	        for(int i = 0; i < n; i++) {
	            int val = x - diff[i];
	            if(end != 0) {
	                int rem = val % end;
	                int quo = val / end;
	                if(rem == 0 && quo >= 0 && (end * quo) == val) {
	                    res++;
	                }
	                //bw.write(val + " " +  end + " " + (val % end) + " " + res +"\n");
	            } else if(end == 0) {
	                if(x == diff[i]) {
	                    res = -1;
	                    break;
	                }
	            }
	        }
	        if(res >= 0) {
	            if(x == 0) {
	                res += 1;
	            }
	        }
	        bw.write(res + "\n");
		}
		bw.flush();
	}
	
}