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.io.DataInputStream;
        import java.io.FileInputStream;
        import java.io.IOException;
        import java.io.InputStreamReader;
        import java.util.*;
        import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args)
    {
        Scanner obj = new Scanner(System.in);
        int t=obj.nextInt();
        for(int q=0;q<t;q++) {
            int n=obj.nextInt(),cur;
            int need=obj.nextInt();
            int[] value=new int[n];
            ArrayList<Integer> l1=new ArrayList<>();
            ArrayList<Integer> l2=new ArrayList<>();

            for(int i=0;i<n;i++){
                value[i]=obj.nextInt();
            }
            int cs=0;
            for(int i=0;i<n;i++){
                cur=obj.nextInt();
                if(cur==1){
                    cs+=value[i];
                    l1.add(value[i]);
                }else{
                    l2.add(value[i]);
                }
            }

            Collections.sort(l1,Collections.reverseOrder());
            Collections.sort(l2,Collections.reverseOrder());
            HashMap<Integer,Integer> hm=new HashMap<>();
            TreeSet<Integer> ts=new TreeSet<>();
            ts.add(Integer.MAX_VALUE);
            int sum=0,temp,low,ans=Integer.MAX_VALUE;
            for(int i=0;i<l2.size();i++) {
                sum += l2.get(i);
                hm.put(sum, i + 1);
                ts.add(sum);
            }
            //cs
            //System.out.println(sum+ " "+cs);
            if(sum+cs<need){
                System.out.println(-1);
                continue;
            }

            //System.out.println(ts);
            //System.out.println(hm+" cs="+cs);
            int f=0;
            if(l1.size()==0) {
                f=1; l1.add(0);
            }
            for (int i = l1.size(); i >= 0; i--) {
                    if (cs < need) {
                        temp = need - cs;
                        //System.out.println("temp="+temp);
                        low=ts.ceiling(temp);
                        if(low==Integer.MAX_VALUE) continue;
                        if(f==1) ans=Math.min(ans,2*hm.get(low));
                        else ans=Math.min(ans,i+2*hm.get(low));
                    }else {
                        //System.out.println("cs="+cs+" i="+i);
                        ans=Math.min(ans,i);
                    }
                    if(i==0) break;
                    cs -= l1.get(i - 1);
                //System.out.println(ans+" "+cs);
            }
            System.out.println(ans);
            //System.out.println();
        }


    }
}
