Polycarp often uses his smartphone. He has already installed n applications on it. Application with number i takes up a_i units of memory.

Polycarp wants to free at least m units of memory (by removing some applications).

Of course, some applications are more important to Polycarp than others. He came up with the following scoring system — he assigned an integer b_i to each application: 

  * b_i = 1 — regular application; 
  * b_i = 2 — important application. 



According to this rating system, his phone has b_1 + b_2 + … + b_n convenience points.

Polycarp believes that if he removes applications with numbers i_1, i_2, …, i_k, then he will free a_{i_1} + a_{i_2} + … + a_{i_k} units of memory and lose b_{i_1} + b_{i_2} + … + b_{i_k} convenience points.

For example, if n=5, m=7, a=[5, 3, 2, 1, 4], b=[2, 1, 1, 2, 1], then Polycarp can uninstall the following application sets (not all options are listed below): 

  * applications with numbers 1, 4 and 5. In this case, it will free a_1+a_4+a_5=10 units of memory and lose b_1+b_4+b_5=5 convenience points; 
  * applications with numbers 1 and 3. In this case, it will free a_1+a_3=7 units of memory and lose b_1+b_3=3 convenience points. 
  * applications with numbers 2 and 5. In this case, it will free a_2+a_5=7 memory units and lose b_2+b_5=2 convenience points. 



Help Polycarp, choose a set of applications, such that if removing them will free at least m units of memory and lose the minimum number of convenience points, or indicate that such a set does not exist.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case contains two integers n and m (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 10^9) — the number of applications on Polycarp's phone and the number of memory units to be freed.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the number of memory units used by applications.

The third line of each test case contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ 2) — the convenience points of each application.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output on a separate line: 

  * -1, if there is no set of applications, removing which will free at least m units of memory; 
  * the minimum number of convenience points that Polycarp will lose if such a set exists. 

Example

Input


5
5 7
5 3 2 1 4
2 1 1 2 1
1 3
2
1
5 10
2 3 2 3 2
1 2 1 2 1
4 10
5 1 3 4
1 2 1 2
4 5
3 2 1 2
2 1 2 1


Output


2
-1
6
4
3

Note

In the first test case, it is optimal to remove applications with numbers 2 and 5, freeing 7 units of memory. b_2+b_5=2.

In the second test case, by removing the only application, Polycarp will be able to clear only 2 of memory units out of the 3 needed.

In the third test case, it is optimal to remove applications with numbers 1, 2, 3 and 4, freeing 10 units of memory. b_1+b_2+b_3+b_4=6.

In the fourth test case, it is optimal to remove applications with numbers 1, 3 and 4, freeing 12 units of memory. b_1+b_3+b_4=4.

In the fifth test case, it is optimal to remove applications with numbers 1 and 2, freeing 5 units of memory. b_1+b_2=3.
import java.util.*;
import java.io.*;

public class Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    //    long mod = 998244353;
    long mod = (long)1e9+7;
    long sum = 0;
    List<Integer>[] tree;
    int[] dp;
    public static void main(String[] args) throws Exception {
        Main main = new Main();
//        int t = Integer.parseInt(reader.readLine());
        int t = sc.nextInt();
//        int t = 1;
        for(int i=1; i<=t; i++){
            main.solve();
        }
        out.flush();

    }
    void solve() throws Exception{
        int n = neInt(), m = neInt();
        ArrayList<Integer> one = new ArrayList<>(), two = new ArrayList<>();
        int[] a = new int[n];
        for(int i=0; i<n; i++){
            a[i] = neInt();
        }
        for(int i=0; i<n; i++){
            int b = neInt();
            if(b==1) one.add(a[i]);
            else two.add(a[i]);
        }
        Collections.sort(one, Collections.reverseOrder());
        Collections.sort(two, Collections.reverseOrder());
        int i1 = one.size()-1;
        long sum1 = 0, sum2 = 0;
        int ans = 3*n;
        for(int i=0; i<one.size(); i++){
            sum1 += one.get(i);
            if(sum1 >= m) ans = Math.min(ans, i+1);
        }
        for(int i=0; i<two.size(); i++){
            sum2 += two.get(i);
            if(sum2 >= m) ans = Math.min(ans, 2*(i+1));
            while(sum2+sum1>=m){
                ans = Math.min(ans, 2*(i+1)+i1+1);
                if(i1>=0){
                    sum1 -= one.get(i1); i1--;
                } else{
                    break;
                }
            }
        }
        if(ans>=3*n) out.println(-1);
        else out.println(ans);
    }
    String neS(){
        return sc.next();
    }
    int neInt(){
        return Integer.parseInt(sc.next());
    }
    long neLong(){
        return Long.parseLong(sc.next());
    }

    int paIn(String s){return Integer.parseInt(s);}


}