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.IOException;
import java.io.InputStreamReader;

public class InfinitePrefixes {
    public static void main(String[] args){
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
            int t = Integer.parseInt(reader.readLine());
            for (int i=0;i<t;i++){
                String[] len_bal = reader.readLine().split(" ");
                int len = Integer.parseInt(len_bal[0]);
                int reqbal = Integer.parseInt(len_bal[1]);

                char[] str = reader.readLine().toCharArray();

                int[] reqcng = new int[len];//remainders

                int abal = 0;
                int ans =0;
                if(reqbal==0){
                    ans = 1;
                }
                for (int j = 0; j < len; j++) {
                    if(str[j] == '0'){
                        abal++;
                    }else abal--;
                    reqcng[j] = reqbal-abal;
                }
                int cngperstr = (reqbal-reqcng[len-1]);
                boolean inf = false;
                if(cngperstr == 0){
                    for (int j = 0; j < len; j++) {
                        if (reqcng[j]==0){
                            inf = true;
                            break;
                        }
                    }
                    if(inf){
                        ans = -1;
                    }
                }else {
                    for (int j = 0; j < len; j++) {
                        if (reqcng[j] / cngperstr >= 0) {
                            if (reqcng[j] % cngperstr == 0) {
                                ans++;
                            }
                        }
                    }
                }
                System.out.println(ans);
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}