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.
// Don't place your source in a package
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;






// Please name your class Main
public class Main {
    static Scanner in = new Scanner(System.in);
    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(System.out);

        int T=Int();
        for(int t=0;t<T;t++){
            int n=Int();
            int m=Int();
            int A[][]=new int[n][2];

            for(int i=0;i<n;i++){
                A[i][0]=Int();
            }

            for(int i=0;i<n;i++){
                A[i][1]=Int();
            }

            Solution sol=new Solution();
            sol.solution(out,A,m);

        }
        out.flush();

    }

    public static long Long(){
        return in.nextLong();
    }
    public static int Int(){
        return in.nextInt();
    }
    public static String Str(){
        return in.next();
    }
}


class Solution{
    List<Integer>one=new ArrayList<>();
    List<Integer>two=new ArrayList<>();
    long pre[];
    public void solution(PrintWriter out,int A[][],int m){
        if(m==0){
            System.out.println(0);
            return;
        }
        int l=1,r=0;
        for(int i=0;i<A.length;i++){
            r+=A[i][1];

            if(A[i][1]==1){
                one.add(A[i][0]);
            }
            else{
                two.add(A[i][0]);
            }
        }

        Collections.sort(one,(a,b)->{
            return b-a;
        });

        Collections.sort(two,(a,b)->{
            return b-a;
        });


        pre=new long[two.size()];
        long sum=0;
        for(int i=0;i<pre.length;i++){
            sum+=two.get(i);
            pre[i]=sum;
        }


        int res=-1;
        while(l<=r){
            int mid=l+(r-l)/2;
            if(check(A,mid,m)){
                res=mid;
                r=mid-1;
            }
            else{
                l=mid+1;
            }
        }
        System.out.println(res);

    }

    public boolean check(int A[][],int mid,int m){
        if(two.size()>0&&((mid/2)-1)>=0&&pre[Math.min((mid/2)-1,pre.length-1)]>=m){
            return true;
        }


        long sum=0;
        for(int i=0;i<one.size();i++){
            sum+=one.get(i);
            mid--;
            if(sum>=m){
                return true;
            }
            if(mid==0)break;


            if(two.size()>0&&((mid/2)-1)>=0&&pre[Math.min((mid/2)-1,pre.length-1)]+sum>=m){
                return true;
            }
        }
        return false;
    }



}

