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.
//package com.company;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
	// write your code here
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while(t>0){
            t--;
            int n = scan.nextInt();
            int m = scan.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            int sum = 0;
            for(int i = 0;i<n;i++){
                a[i] = scan.nextInt();
                sum += a[i];
            }
            for(int i = 0;i<n;i++){
                b[i] = scan.nextInt();
            }

            
                ArrayList<Integer> norm = new ArrayList<>();
                ArrayList<Integer> imp = new ArrayList<>();
                for(int i = 0;i<n;i++){
                    if(b[i] == 2){
                        imp.add(a[i]);
                    }
                    else{
                        norm.add(a[i]);
                    }
                }
                Collections.sort(norm, Collections.reverseOrder());
                Collections.sort(imp, Collections.reverseOrder());
//                System.out.println(norm);
                long ans = Long.MAX_VALUE;
                long[] sum_norm = new long[norm.size() + 1];
                long[] sum_imp = new long[imp.size() + 1];
//                Arrays.fill(sum_imp,0);
//                Arrays.fill(sum_norm,0);
//                System.out.println(Arrays.toString(sum_norm));
                for(int i = 0;i<norm.size();i++){
//                    System.out.println(i);
//                    System.out.println(Arrays.toString(sum_norm));
                    sum_norm[i + 1] = sum_norm[i] + norm.get(i);
                }
                for(int i = 0;i<imp.size();i++){
                    sum_imp[i + 1] = sum_imp[i] + imp.get(i);
                }
                for(int i = 0;i<norm.size()+1;i++){
                    int l = 0, r = imp.size();
                    long pts = i;
                    long rem = (m - sum_norm[i]);
                    while(l < r){
                        int mid = (l + r) / 2;
                        if(sum_imp[mid] < rem){
                            l = mid+1;
                        }
                        else{
                            r = mid;
                        }
                    }
                    if(sum_imp[l] >= rem ){
                        ans = Math.min(ans, pts + l*2);
                    }
                }
                System.out.println(ans == Long.MAX_VALUE ? -1 : ans);
            

        }
    }
}
