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 class Pair implements Comparable<Pair> {
        int a;
        int b;
        Pair(int a,int b) {
            this.a=a;
            this.b=b;
        }
       public   int compareTo(Pair p) {
         if(this.b==p.b) {
             if(this.b==1)
             return this.a-p.a;
             else 
             return p.a-this.a;
         }
         else 
         return this.b-p.b;
        }
    }
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0) {
            int n=sc.nextInt();
            int m=sc.nextInt();
            int a[]=new int[n];
            int b[]=new int[n];
            Pair pa[]=new Pair[n];
            long sum=0;
            for(int i=0;i<n;i++) {
                a[i]=sc.nextInt();
                sum+=a[i];
            }
            
             for(int i=0;i<n;i++) {
                b[i]=sc.nextInt();
            }
            if(sum<m) {
                System.out.println("-1");
            }
            else {
            for(int i=0;i<n;i++)
             pa[i]=new Pair(a[i],b[i]);
             
            //  for(int i=0;i<n;i++)
            //  System.out.println(pa[i].a+" "+pa[i].b);
             
             
             Arrays.sort(pa);
             int left=0;
             int ans=Integer.MAX_VALUE;
             int cost=0;
             long curr=0;
             for(int i=0;i<n;i++) {
                 curr+=pa[i].a;
                 cost+=pa[i].b;
                 while(curr>=m) {
                     ans=Math.min(cost,ans);
                     curr-=pa[left].a;
                     cost-=pa[left].b;
                     left++;
                 }
             }
             System.out.println(ans);
            }
        }

       


        }
}
