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 Solution{
    public static void main(String arg[]){
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while (t > 0)
        {
		    int i, j, n = scan.nextInt();
 			long k = scan.nextLong(), a = 0;
			long [] arr = new long[n];
			for (i = 0; i < n; ++i)
			{
			    arr[i] = scan.nextLong();
				a += arr[i];
			}
			ArrayList <Long> arr1 = new ArrayList<Long>();
            ArrayList <Long> arr2 = new ArrayList<Long>();
	        for (i = 0; i < n; ++i)
            {
                j = scan.nextInt();
                if (j == 1)
                {
                arr1.add(arr[i]);
                }
                else
                {
                arr2.add(arr[i]);
                } 				
 			}
 			if (a >= k)
 			{
            Collections.sort(arr1, Collections.reverseOrder());
			Collections.sort(arr2, Collections.reverseOrder());
			long [] arr3 = new long[arr2.size() + 1];
			arr3[0] = 0;
			for (i = 1; i <= arr2.size(); ++i)
			{
			arr3[i] += arr3[i - 1] + arr2.get(i - 1);
			}
			long p = 0, q = 0, r = Integer.MAX_VALUE, x, y;
		    j = Arrays.binarySearch(arr3, k);
			if (j < 0)
			{
			j++;
			j *= -1;
			if (j < arr3.length)
			{
			if (arr3[j] >= k)
			{
			r = Math.min(r, 2L*j);
			}
			}
			}
			else
			{
			r = Math.min(r, 2L*j);
			}
			for (i = 0; i < arr1.size(); ++i)
			{
			p += arr1.get(i);
			q++;
			if (p >= k)
			{
			r = Math.min(r, q);
			}
			else
			{
			j = Arrays.binarySearch(arr3, (k - p));
			if (j < 0)
			{
			j++;
			j *= -1;
			if (j < arr3.length)
			{
			if (arr3[j] + p >= k)
			{
			r = Math.min(r, 2L*j + q);
			}
			}
			}
			else
			{
			r = Math.min(r, 2L*j + q);
			}
			}
			}
			System.out.println(r);
			}
			else
			{
			   System.out.println("-1"); 
			}
            t--;
        }
    }
}