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.
//  package com.company;



import com.sun.security.jgss.GSSUtil;

import javax.swing.plaf.IconUIResource;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.lang.management.MonitorInfo;
import java.nio.file.LinkOption;
import java.sql.SQLOutput;
import java.util.*;

//b

public class Main {
    public static void main(String[] args) {
        Scanner input=new Scanner(System.in);
        int t=input.nextInt();
        while (t-->0){
           int n=input.nextInt();
           long m=input.nextLong();
           long a[]=new long[n];
           ArrayList<Long> one=new ArrayList<>();
           ArrayList<Long> two=new ArrayList<>();
            for (int i = 0; i <n ; i++) {
                a[i]=input.nextLong();
            }
            for (int i = 0; i <n ; i++) {
                int curr=input.nextInt();
                if (curr==1) one.add(a[i]);
                else two.add(a[i]);
            }
            TreeMap<Long,Long> map=new TreeMap<>();
            two.sort(Collections.reverseOrder());
            long p=0;
            long c=0;
            for (int i = 0; i<two.size() ; i++) {
                p+=two.get(i);
                c+=2;
                map.put(p,c);
            }
            long ans=Integer.MAX_VALUE;
            if (  map.ceilingKey(m)!=null) ans=map.get(map.ceilingKey(m));
            p=0;
            one.sort(Collections.reverseOrder());
            for (int i = 0; i <one.size(); i++) {
                p+=one.get(i);
              long req=m-p;
              if (req>0 ){
                  Long twocost=map.ceilingKey(req);
                  if (twocost!=null){
                      ans=Math.min(ans,i+1+map.get(twocost));
                  }
              }else {
                  ans=Math.min(ans,i+1);
                  break;
              }
            }

            if (ans==Integer.MAX_VALUE){
                System.out.println(-1);
                continue;
            }
            System.out.println(ans);

        }

    }
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }



}
//8 -1 9 -1 8 8 -1 -1 -1 8
// 9 -1 1 2 -1 2 9 -1 -1
