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.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Solve5 {

    public static void main(String[] args) throws IOException {
        PrintWriter pw = new PrintWriter(System.out);
        new Solve5().solve(pw);
        pw.flush();
        pw.close();
    }

    public void solve(PrintWriter pw) throws IOException {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt(), x = sc.nextInt();
            char[] s = sc.next().toCharArray();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                if (s[i] == '0') {
                    a[i] = (i > 0 ? a[i - 1] : 0) + 1;
                } else {
                    a[i] = (i > 0 ? a[i - 1] : 0) - 1;
                }
            }
            int ans = (x == 0 ? 1 : 0);
            int inc = a[n - 1];
            for (int i = 0; i < n; i++) {
                if (x - a[i] <= 0 && inc < 0 && (x - a[i]) % inc == 0) {
                    ++ans;
                } else if (x - a[i] >= 0 && inc > 0 && (x - a[i]) % inc == 0) {
                    ++ans;
                } else if (x - a[i] == 0 && inc  == 0) {
                    ++ans;
                }
            }
            if (inc == 0 && ans > 0) {
                pw.println(-1);
            } else {
                pw.println(ans);
            }
        }
    }

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

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

        public FastReader(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
            }
        }

        public String next() {
            if (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                }
            }
            return st.nextToken();
        }

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

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

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

        public String nextLine() {
            try {
                return br.readLine();
            } catch (Exception e) {
            }
            return null;
        }

        public boolean hasNext() throws IOException {
            if (st != null && st.hasMoreTokens()) {
                return true;
            }
            String s = br.readLine();
            if (s == null || s.isEmpty()) {
                return false;
            }
            st = new StringTokenizer(s);
            return true;
        }
    }
}