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.*;
public class prac {
  static List<Integer> a1,a2;
  static int m1,l1,l2;
    public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0)
        {
            int n=sc.nextInt(),m=sc.nextInt();
            m1=m;
            int[] a=new int[n],b=new int[n];
            a1=new ArrayList<Integer>();
            a2=new ArrayList<Integer>();
            for(int i=0;i<n;i++)
            {
            a[i]=sc.nextInt();
            }
            for(int i=0;i<n;i++)
            {
            b[i]=sc.nextInt();
            }
            for(int i=0;i<n;i++)
            {
              if(b[i]==1)
              a1.add(a[i]);
              else
              a2.add(a[i]);
            }
            l1=a1.size();
            l2=a2.size();
            Collections.sort(a1, Collections.reverseOrder());
            Collections.sort(a2, Collections.reverseOrder());
            long sum=0;
            for(int i:a)
            sum+=i;
            if(sum<m)
            {
              System.out.println("-1");
              continue;
            }
            int sum1=0;
            for(int i:b)
            sum1+=i;
            int ans=Integer.MAX_VALUE;
            int start=0,end=sum1;
            int mid=0;
            while(start<=end)
            {
             mid=start+(end-start)/2;
              if(isPossible(mid))
              {
                 end=mid-1;
                 ans=Math.min(ans,mid);
              }
              else
              start=mid+1;
            }
            System.out.println(start);
            

        }
      }
      static Map<Integer,Integer> map;
      public static boolean isPossible(int n)
      {
        
         map=new HashMap<>();
         util(n, 0, 0, 0);
         for(Map.Entry<Integer,Integer> entry:map.entrySet())
         {
           int sum=0;
           int i=entry.getKey();
           int j=entry.getValue();
          //  System.out.println(i+" "+j+"--->"+n);
           if(i>l1 || j>l2)
           continue;
           for(int k=0;k<i;k++)
           {
            sum+=a1.get(k);
           }
           for(int k=0;k<j;k++)
           {
            sum+=a2.get(k);
           }
           if(sum>=m1)
           return true;
         }
         return false;
      }
      public static void util(int n,int one,int two,int te)
      {
        if(te>n)
        return;
        if(te==n)
        map.put(one, two);
        util(n, one+1,two, te+1);
        util(n, one, two+1, te+2);
      }
     }        

