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.training;

import javafx.util.Pair;

import javax.swing.*;
import java.util.*;

public class Main {

    public static void main(String[] args) {
        // write your code here

        Scanner scanner = new Scanner(System.in);
        int cases = scanner.nextInt();

        for (int i = 0; i < cases; i++)
        {
            int a = scanner.nextInt();
            int b = scanner.nextInt();


            int[] boys = new int[a];
            for (int j = 0; j < a; j++)
                boys[j] = scanner.nextInt();

            int[] girls = new int[a];
            for (int j = 0; j < a; j++)
                girls[j] = scanner.nextInt();


            System.out.println(solve(a,b,boys,girls));

        }
    }

    //public static boolean[] dp = new boolean[1000001];

    public static long solve(int a, int b, int[] boys, int[] girls)
    {


        List<Integer> one = new ArrayList<>();
        List<Integer> two = new ArrayList<>();




        for (int i = 0; i < a; i++)
        {
            if(girls[i] == 1)
                one.add(boys[i]);
            else
                two.add(boys[i]);
        }

        one.sort(Collections.reverseOrder());
        two.sort(Collections.reverseOrder());

        int k = b;

        int points = 0;
        int i = 0;
        int j = 0;
        int ones = one.size();
        int twos = two.size();
        while (k > 0 && ones>1 && twos>0)
        {
             if(k <= one.get(i))
                return points + 1;

             if(k <= one.get(i)+one.get(i+1))
                 return points + 2;

            if(k <= two.get(j))
                return points + 2;

            if(k <= one.get(i)+two.get(j))
                return points + 3;

             if(one.get(i)+one.get(i+1) > two.get(j))
                {
                    k -= one.get(i);
                    ones -= 1;
                    i ++;
                    points++;

                }else
                {
                    twos --;
                    k -= two.get(j);
                    j++;
                    points += 2;
                }



        }

        if(k <= 0)
            return points;
        if(ones <= 1){
            while (k > 0 && twos > 0)
            {
                if(ones == 1 && k <= one.get(i))
                    return points + 1;

                if(k <= two.get(j))
                    return points + 2;

                if(ones == 1 && k <= one.get(i) + two.get(j))
                    return points + 3;

                twos --;
                k -= two.get(j);
                j++;
                points += 2;
            }
        }else
        {
            while (k > 0 && ones > 0)
            {
                k -= one.get(i);
                i++;
                points++;
                ones--;
            }
        }



        return k > 0 ? -1 : points;


    }


}
