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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;



public final class Solution {
	
	static class in {
	    static BufferedReader reader;
	    static StringTokenizer tokenizer;

	    /** call this method to initialize reader for InputStream */
	    static void init(InputStream input) {
	        reader = new BufferedReader(
	                     new InputStreamReader(input) );
	        tokenizer = new StringTokenizer("");
	    }

	    /** get next word */
	    static String next() throws IOException {
	        while ( ! tokenizer.hasMoreTokens() ) {
	            //TODO add check for eof if necessary
	            tokenizer = new StringTokenizer(
	                   reader.readLine() );
	        }
	        return tokenizer.nextToken();
	    }

	    static int nextInt() throws IOException {
	        return Integer.parseInt( next() );
	    }
		
	    float nextFloat() throws IOException {
	        return Float.parseFloat( next() );
	    }

		public static long nextLong() throws NumberFormatException, IOException {
			// TODO Auto-generated method stub
			return Long.parseLong( next() );
		}
	}
	
public static final void main(String[] args) throws IOException{
 in.init(System.in);
 int t=in.nextInt();

   for(int p=0;p<t;p++)
 {   int n=in.nextInt();
 	 long m=in.nextLong();
 	 ArrayList<Long> b1=new ArrayList();
 	 ArrayList<Long> b2=new ArrayList();
 	 long[] arr=new long[n];
 	 for(int i=0;i<n;i++)
 	 {
 		 arr[i]=in.nextLong();
 	 }
 	 
 	for(int i=0;i<n;i++)
	 {
		 if(in.nextInt()==1)
			 b1.add(arr[i]);
		 else
			 b2.add(arr[i]);
		 
	 }
 	Collections.sort(b1,Comparator.reverseOrder());
 	Collections.sort(b2,Comparator.reverseOrder());
 	long[] prefb1=new long[b1.size()];
 	long[] prefb2=new long[b2.size()];
 	long temp=0;
 	for(int i=0;i<b1.size();i++)
 	{
 		temp+=b1.get(i);
 		prefb1[i]=temp;
 	}
 	temp=0;
 	for(int i=0;i<b2.size();i++)
 	{
 		temp+=b2.get(i);
 		prefb2[i]=temp;
 	}
 	
 	long ans=Long.MAX_VALUE;
 	if(m>(b1.size()>0?prefb1[b1.size()-1]:0)+(b2.size()>0?prefb2[b2.size()-1]:0))
 		ans=-1;
 	else
 	{
	 	for(int i=b1.size()-1;i>=0;i--)
	 	{
	 		long sol=(i+1);
	 		long left=m-prefb1[i];
	 		//System.out.println(left+" "+sol);
	 		if(left<=0)ans=Math.min(ans,sol);
	 		else
	 		{int pos=Arrays.binarySearch(prefb2,left);
	 	//	System.out.println("left:"+left);
	 		if(pos>=0)
	 		{
	 			sol+=(pos+1)*2;
	 			ans=Math.min(sol,ans);
	 		}
	 		else
	 		{
	 		//	System.out.println("neg"+pos);
	 			pos=-(pos+1);
	 		
	 			if(pos>=b2.size())break;
			//System.out.println("here:"+prefb2[pos]);
	 			//System.out.println(pos);
	 			sol+=(pos+1)*2;
	 			//System.out.println("sol:"+sol);
	 			ans=Math.min(sol,ans);
	 			
	 		}
	 		}
	 	}
	 	
	 	int pos=Arrays.binarySearch(prefb2,m);
	 	long sol=0;
	 	if(pos>=0)
 		{
 			sol+=(pos+1)*2;
 			ans=Math.min(sol,ans);
 		}
 		else
 		{
 			pos=-(pos+1);
 			if(pos<b2.size())
 			{sol+=(pos+1)*2;
 			ans=Math.min(sol,ans);}
 		}
	 
 	}
 	System.out.println(ans);
	
 }
}
}