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 ProblemD {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();

		for (int i = 0; i < t; i++) {
			int n, m;
			n = s.nextInt();
			m = s.nextInt();
			int input[][] = new int[n][2];
			for (int j = 0; j < n; j++) {
				input[j][0] = s.nextInt();
			}
			for (int j = 0; j < n; j++) {
				input[j][1] = s.nextInt();
			}
			ArrayList<Integer> important = new ArrayList<>();
			ArrayList<Integer> unimportant = new ArrayList<>();
			for (int j = 0; j < n; j++) {
				if (input[j][1] == 1) {
					unimportant.add(input[j][0]);

				} else {
					important.add(input[j][0]);
				}
			}
			Collections.sort(important);
			Collections.sort(unimportant);
			TreeSet<Long> unimportant_sum = new TreeSet<>();
			TreeSet<Long> important_sum = new TreeSet<>();
			TreeMap<Long, Integer> important_sum_list = new TreeMap<>();
			long sum = 0;
			unimportant_sum.add((long) 0);
			important_sum.add((long) 0);
			important_sum_list.put((long) 0, 0);
			for (int j = unimportant.size() - 1; j >= 0; j--) {
				sum += (long) unimportant.get(j);
				unimportant_sum.add(sum);
			}
			sum = 0;
			for (int j = important.size() - 1; j >= 0; j--) {
				sum += (long) important.get(j);
				important_sum.add(sum);
				important_sum_list.put(sum, (important.size() - j) * 2);
			}

			int ans = (int) 1e9;
			int index = -1;
			for (Long l : unimportant_sum) {

				Integer target;
				index++;

				if (important_sum.ceiling(m - l) == null) {

					continue;
				}

				target = important_sum_list.get(important_sum.ceiling(m - l));
				
				ans = Math.min(index + target, ans);

			}

			if (ans == 1e9) {
				System.out.println(-1);
			} else {
				System.out.println(ans);
			}

		}
	}

}
