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.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
  
    void solve() throws IOException {
        outer:
        for (int TC=ni();TC>0;TC--) {
            int n=ni(),p=ni();
            char[]A= next().toCharArray();
            int b=0;
            for (int x=0;x<n;x++) {
                if (A[x]=='1') b--;
                else b++;
            }
            
            if (b==0) {
                if (p==0) out.println(-1);
                else {
                    int c=0;
                    for (int x=0;x<n;x++) {
                        if (A[x]=='1') c--;
                        else c++;
                        if (c==p) {out.println(-1); continue outer; }
                    }
                out.println(0);
                }
                continue outer;
            }
            
            long ans=0;
            if (b>0) {
                if (p==0) ans++;
                int c=0;
                for (int x=0;x<n;x++) {
                    if (A[x]=='1') c--;
                    else c++;
                    if (p>=c && (p-c)%b==0) ans++;
                }
            }
            else {
                if (p==0) ans++;
                int c=0;
                for (int x=0;x<n;x++) {
                    if (A[x]=='1') c--;
                    else c++;
                    if (p<=c && (p-c)%b==0) ans++;
                }
            }
            
            out.println(ans);
        }
        out.flush();
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}