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:
// package EducationalRound81;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;

public class ProblemB {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int test=Integer.parseInt(br.readLine());
        while(test--!=0){
            String line[]=br.readLine().split(" ");
            int n=Integer.parseInt(line[0]);
            int x=Integer.parseInt(line[1]);
            String str=br.readLine();
            int a[]=new int[n+1];
            int t=Math.abs(x);
            if(x>=0){
                HashMap<Integer,Integer> count=new HashMap<>();
                for(int i=1;i<=n;i++){
                    char c=str.charAt(i-1);
                    if(c=='0'){
                        a[i]=a[i-1]+1;
                    }
                    else{
                        a[i]=a[i-1]-1;
                    }
                    count.put(a[i],count.getOrDefault(a[i],0)+1);
                }
                if(a[n]==0&&count.containsKey(t)){
                    System.out.println(-1);
                    continue;
                }
                if(a[n]==0&&!count.containsKey(t)){
                    System.out.println(0);
                    continue;
                }
                int ans=0;
                for(int k:count.keySet()){
                    int temp=(t-k)/a[n];
//                    System.out.println(temp);
                    if((t-k)%a[n]==0&&temp>=0){
                        ans+=count.get(k);
                    }
                }
                if(x==0){
                    ans++;
                }
                System.out.println(ans);
            }
            else{
                HashMap<Integer,Integer> count=new HashMap<>();
                for(int i=1;i<=n;i++){
                    char c=str.charAt(i-1);
                    if(c=='1'){
                        a[i]=a[i-1]+1;
                    }
                    else{
                        a[i]=a[i-1]-1;
                    }
                    count.put(a[i],count.getOrDefault(a[i],0)+1);
                }
                if(a[n]==0&&count.containsKey(t)){
                    System.out.println(-1);
                    continue;
                }
                if(a[n]==0&&!count.containsKey(t)){
                    System.out.println(0);
                    continue;
                }
                int ans=0;
                for(int k:count.keySet()){
                    int temp=(t-k)/a[n];
//                    System.out.println(temp);
                    if((t-k)%a[n]==0&&temp>=0){
                        ans+=count.get(k);
                    }
                }
                System.out.println(ans);
            }
        }
    }
}