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.*;
import java.math.BigInteger;
import java.util.*;

public class B {
	static BufferedReader br;
	static int cin() throws Exception
	{
		return Integer.valueOf(br.readLine());
	}
	static int[] split() throws Exception
	{
		String[] cmd=br.readLine().split(" ");
		int[] ans=new int[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Integer.valueOf(cmd[i]);
		}
		return ans;
	}
	static long[] splitL() throws IOException
	{
		String[] cmd=br.readLine().split(" ");
		long[] ans=new long[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Long.valueOf(cmd[i]);
		}
		return ans;
	}
	static long power(long x,long y,long p)
	{
		long res=1;
		x=x%p;
		while(y>0)
		{
			if((y&1)==1)
				res=(res*x)%p;
			y=(y>>1);
			x=(x*x)%p;
		}
		return res;
	}
	static long inv(long x,long p)
	{
		return power(x,p-2,p);
	}
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
		br=new BufferedReader(new InputStreamReader(System.in));
		int t=cin();
		while(t!=0)
		{
			t--;
			int[]ar=split();
			int n=ar[0];
			int m=ar[1];
			int[]a=split();
			int[]b=split();
			ArrayList<Integer>one=new ArrayList<>();
			ArrayList<Integer>two=new ArrayList<>();
			long tot=0;
			for(int i=0;i<n;i++)
			{
				tot+=a[i];
				if(b[i]==1)
					one.add(a[i]);
				else
					two.add(a[i]);
			}
			if(tot<m)
			{
				System.out.println(-1);
				continue;
			}
			Collections.sort(one,Collections.reverseOrder());
			Collections.sort(two,Collections.reverseOrder());
//			System.out.println(one);
//			System.out.println(two);
			long[]pre=new long[two.size()];
			long s=0;
			for(int i=0;i<two.size();i++)
			{
				s+=two.get(i);
				pre[i]=s;
			}
			long ans=(long)1e18;
			s=0;
			for(int i=0;i<=one.size();i++)
			{
				if(i!=0)
					s=s+one.get(i-1);
				long req=(long)m-s;
				if(req<=0)
				{
					ans=Math.min(ans, i);
					continue;
				}
				int ind=Arrays.binarySearch(pre,req);
				//System.out.println(s+" "+req+" "+ind);
				if(ind<0)
					ind=Math.abs(ind);
				else
					ind++;
				if(ind>two.size())
					continue;
				ans=Math.min(ans,i+ind*2);
			}
			System.out.println(ans);
		}
	}
}
