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 whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner scn=new Scanner(System.in);
		int t=scn.nextInt();
		while(t-->0){
			int n=scn.nextInt();
			long m=scn.nextLong();
			long[] arr=new long[n];
			for(int i=0; i<n; i++){
				arr[i]=scn.nextLong();
			}
			ArrayList<Long> type1=new ArrayList<>();
			ArrayList<Long> type2=new ArrayList<>();
			for(int i=0; i<n; i++){
				int x=scn.nextInt();
				if(x==1){
					type1.add(arr[i]);
				}
				else{
					type2.add(arr[i]);
				}
			}
			
			Collections.sort(type1);
			Collections.reverse(type1);
			Collections.sort(type2);
			Collections.reverse(type2);
			int ans=Integer.MAX_VALUE;
			int size1=type1.size();
			int size2=type2.size();
			long[] prefix1=new long[size1+1];
			prefix1[0]=0;
			for(int i=0; i<size1; i++){
				prefix1[i+1]=prefix1[i]+type1.get(i);
			}
			long[] prefix2=new long[size2+1];
			prefix2[0]=0;
			for(int i=0; i<size2; i++){
				prefix2[i+1]=prefix2[i]+type2.get(i);
			}
                  int index=0;
			for(int i=0; i<=size1; i++){
				long remaining=m-prefix1[i];

				if(remaining<=0){
					ans=Math.min(ans,i);
				}
                        if(size2>0){  
				index=lower_bound(prefix2,size2+1, remaining);
                        }
				remaining-=prefix2[index];
				if(remaining<=0){
					ans=Math.min(ans,i+2*index);
				}
				
			}
                        if(ans==Integer.MAX_VALUE){
                              System.out.println(-1);
                        }
                        else{
			            System.out.println(ans);
                        }

			
		}
		
	}
	public static int lower_bound(long[] arr, int N, long X)
      {
            int mid=0;
  
    // Initialise starting index and
    // ending index
            int low = 0;
            int high = N;
      if(X==0){
        return 0;
      }
      if(X<arr[1]){
        return 1;
      }
      if(X>=arr[N-1]){
        return N-1;
      }
    // Till low is less than high
      while (low < high) {
            mid = low + (high - low) / 2;
  
        // If X is less than or equal
        // to arr[mid], then find in
        // left subarray
            if (X <= arr[mid]) {
                  high = mid;
            }
  
        // If X is greater arr[mid]
        // then find in right subarray
            else {
                low = mid + 1;
        }
    }
  
    // Return the lower_bound index
    return low;
      }
}