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 Main {
       
       // Pair class 
    public static class Pair { 
        int a; 
        int b; 
      
        // Constructor 
        public Pair(int a, int b) 
        { 
            this.a = a; 
            this.b = b; 
        } 
    } 
    
    
    public static class Compare { 
        
        public void compare(Pair arr[], int n) 
        { 
            // Comparator to sort the pair according to first element 
            Arrays.sort(arr, new Comparator<Pair>() { 
                @Override public int compare(Pair p1, Pair p2) 
                { 
                    return p1.a - p2.a; 
                } 
            }); 
            
        } 
        
    }

     public static void main(String []args){
        Scanner scn = new Scanner(System.in);
        int t = scn.nextInt();
        
        for(int i = 0; i < t; i++){
            
            int n = scn.nextInt();
            int m = scn.nextInt();
            
            int[] a = new int[n];
            int[] b = new int[n];
            
            for(int j = 0; j < n; j++){
                a[j] = scn.nextInt();
            }
            
            for(int j = 0; j < n; j++){
                b[j] = scn.nextInt();
            }
            
            Pair[] arr = new Pair[n];
            
            for(int j = 0; j < n; j++){
                arr[j] = new Pair(a[j], b[j]);
            }
            
            Compare obj = new Compare(); 
  
            obj.compare(arr, n);
            
            //cummulative array of pair
            for(int j = 1; j < n; j++){
                arr[j].a += arr[j - 1].a;
                arr[j].b += arr[j - 1].b;
            }
            
            //check
            int res = Integer.MAX_VALUE;
            for(int j = 1; j < n; j++) {
                for(int k = j - 1; k >= 0; k--){
                    if(arr[j].a - arr[k].a >= m){
                        if(arr[j].b - arr[k].b < res){
                            res = arr[j].b - arr[k].b;
                        }
                        break;
                    }
                }
            }
            
            if(res == Integer.MAX_VALUE){
                res = -1;
            }
            
            System.out.println(res);
            
        }
     }
}