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.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class InfinitePrefixes {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }

    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt(), x = in.nextInt();
        char[] cs = in.next().toCharArray();
        int cnt = 0;
        Map<Integer, Integer> mp = new HashMap<>();
        int zero = 0, one = 0;
//        mp.put(0, 0);
        for(char c : cs){
            if (c == '0'){
                zero++;
            }else{
                one++;
            }
//            System.out.println((zero - one )+" ffff"+mp);
            int get = zero - one;
            mp.put(get, mp.getOrDefault(get, 0) + 1);
//            System.out.println(mp);
        }
        int dif = zero - one;
        long ans = 0;
        if (dif == 0){
            if (mp.containsKey(x)){
                out.println(-1);
            }else{
                out.println(0);
            }
        }else{
            if (x == 0) ans++;
            for(int k : mp.keySet()){
                int add = mp.get(k);
                if ((x - k) % dif == 0){
                    int req = (x - k) / dif;
                    if (req >= 0){
                        ans += add;
                    }
                }
            }
            out.println(ans);
        }
    }
}