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.*;
import java.util.StringTokenizer;

public class A {

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int tc = sc.nextInt();
        while (tc --> 0) {
            int n = sc.nextInt();
            int x = sc.nextInt();
            char[] a = sc.next().toCharArray();
            int[] preSum = new int[n];
            preSum[0] = a[0] - '0' == 0 ? 1 : -1;
            for (int i = 1; i < n; i++) {
                preSum[i] = preSum[i - 1] + (a[i] - '0' == 0 ? 1 : -1);
            }
            int s = preSum[n - 1];
            int count = x == 0 ? 1 : 0;
            for (int i = 0; i < n; i++) {
                if (s == 0) {
                    if (preSum[i] == x) {
                        count = -1;
                        break;
                    }
                } else {
                    count += (x - preSum[i]) % s == 0 && (x - preSum[i]) / s >= 0 ? 1 : 0;
                }
            }
            out.println(count);
        }
        out.flush();
        out.close();
    }

    static class Scanner
    {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream system) {br = new BufferedReader(new InputStreamReader(system));}
        public Scanner(String file) throws Exception {br = new BufferedReader(new FileReader(file));}
        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
        public String nextLine()throws IOException{return br.readLine();}
        public int nextInt() throws IOException {return Integer.parseInt(next());}
        public double nextDouble() throws IOException {return Double.parseDouble(next());}
        public char nextChar()throws IOException{return next().charAt(0);}
        public Long nextLong()throws IOException{return Long.parseLong(next());}
        public boolean ready() throws IOException{return br.ready();}
        public void waitForInput() throws InterruptedException {Thread.sleep(3000);}
    }
}