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.util.*;
import java.io.*;
public class Main
{
	public static void main(String[] args)throws Exception
	{
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw=new PrintWriter(System.out);
	    int t=Integer.parseInt(br.readLine());
	    while(t-->0)
	    {
	        String s[]=(br.readLine()).split(" ");
	        int n=Integer.parseInt(s[0]);
	        long m=Long.parseLong(s[1]);
	        String s1[]=(br.readLine()).split(" ");
	        String s2[]=(br.readLine()).split(" ");
	        long a[]=new long[n];
	        long b[]=new long[n];
	        LinkedList<Long> one=new LinkedList<Long>();
	        LinkedList<Long> two=new LinkedList<Long>();
	        long sum=0l;
	        for(int i=0;i<n;i++)
	        {
	            a[i]=Long.parseLong(s1[i]);
	            b[i]=Long.parseLong(s2[i]);
	            if(b[i]==1)
	            {
	                one.add(a[i]);
	            }
	            else
	            {
	                two.add(a[i]);
	            }
	            sum+=a[i];
	        }
	        if(sum<m)
	        {
	            pw.println(-1);
	            continue;
	        }
	        long ar[]=new long[n];
	        int l1=one.size();
	        int l2=two.size();
	        Collections.sort(one);
	        Collections.sort(two);
	        Iterator<Long> it1=one.iterator();
	        Iterator<Long> it2=two.iterator();
	        for(int i=l1-1;i>=0;i--)
	        {
	            ar[i]=(long)(it1.next());
	        }
	        for(int i=l1;i<n;i++)
	        {
	            ar[i]=(long)(it2.next());
	        }
	        sum=m;
	        int ct=0;
	        int l=0;int r=n-1;
	        while(sum>0)
	        {
	            if(l>=l1)
	            {
	                sum-=ar[r];
	                r-=1;
	                ct+=2;
	            }
	            else if(r<l1)
	            {
	                sum-=ar[l];
	                l++;
	                ct+=1;
	                if(sum>0)
	                {
	                    sum-=ar[l];
	                    l++;
	                    ct+=1;
	                }
	            }
	            else
	            {
	                if(sum<=ar[l])
	                {
	                    sum-=ar[l];
	                    l++;
	                    ct+=1;
	                    break;
	                }
	                
	                //else if((l+1)>=l1 && ar[l]>=ar[r])
	                //testcomment
	                else if(ar[l]>=ar[r])//also handles the case where just one element on l side is left
	                {
	                    sum-=ar[l];
	                    l++;
	                    ct+=1;
	                    
	                }
	                else if((l+1)<l1 && (ar[l]+ar[l+1])>ar[r])
	                {
	                    if(sum<=(ar[l]+ar[l+1]))
	                    {
	                        sum-=(ar[l]+ar[l+1]);
	                        l+=2;
	                        ct+=2;
	                        break;
	                    }
	                    else
	                    {
	                        if((l+2)<l1 )// || (ar[l+1]+ar[l])<(ar[l]+ar[r]))
	                        {
	                            if((ar[l]+ar[l+1]+ar[l+2])<=(ar[l]+ar[r]))
	                            {
    	                            sum-=ar[l];
	                                sum-=ar[r];
	                                l++;
	                                r--;
	                                ct+=3;
	                            }
	                            else
	                            {
	                                sum-=(ar[l]+ar[l+1]+ar[l+2]);
	                                l+=3;
	                                ct+=3;
	                            }
	                        }
	                        else if(((ar[l+1]+ar[l])<(ar[l]+ar[r])) &&((ar[l]+ar[r])>=sum))
	                        {
	                                sum-=ar[l];
	                                sum-=ar[r];
	                                l++;
	                                r--;
	                                ct+=3;
	                        }
	                        else
	                        {
	                            sum-=(ar[l]+ar[l+1]);
	                            l+=2;
	                            ct+=2;
	                            
	                        }
	                        
	                    }
	                    
	                    
	                }
	                else
	                {
	                    sum-=ar[r];
	                    r--;
	                    ct+=2;
	                }
	            }
	            
	        }
	        pw.println(ct);
	    }
	    pw.flush();
	}
}
