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.awt.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
import java.util.List;

public class Main {
    static int mod = (int) 1e9 + 7;
    public static void main(String[] args){
        FastReader sc = new FastReader();
        StringBuilder sb=new StringBuilder();
        int t=sc.nextInt();
        while(t-->0) {
            int n = sc.nextInt(), x = sc.nextInt();
            String s = sc.nextLine();
            int[] zero = new int[n];
            zero[0] = s.charAt(0) == '0' ? 1 : 0;
            for (int i = 1; i < n; i++) {
                zero[i] = zero[i - 1] + (s.charAt(i) == '0' ? 1 : 0);
            }
            int[] bal = new int[n];
            for (int i = 0; i < n; i++) {
                bal[i] = zero[i] - (i + 1 - zero[i]);
            }
            if (bal[n - 1] == 0) {
                boolean z = false;
                for (int i = 0; i < n; i++) {
                    if (bal[i] == x) {
                        z = true;
                        break;
                    }
                }
                if (z)sb.append(-1+"\n");
                else sb.append(0+"\n");
            } else {
                int ans = 0, temp;
                for (int i = 0; i < n; i++) {
                    temp = x - bal[i];
                    if(temp%bal[n-1]==0 && temp/bal[n-1]>=0)ans++;
                }
                if(x==0)ans++;
                sb.append(ans+"\n");
            }
        }
        System.out.println(sb);
    }
}

class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}