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.io.*;
import java.util.*;
import java.util.stream.*;


public class CleaningThePhone {

	public static String getLine(BufferedReader bufferedReader) throws Exception {
		return bufferedReader.readLine().trim();
	}

	public static void main(String[] args) throws Exception {
		BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(getLine(bufferedReader));
		
		for (int t = 0; t < T; t++) {
		
			
			List<Integer> NM = Arrays.stream(getLine(bufferedReader).split(" "))
				.map(Integer::parseInt)
				.collect(Collectors.toList());
			int N = NM.get(0);
			int M = NM.get(1);
			List<Integer> As = Arrays.stream(getLine(bufferedReader).split(" "))
				.map(Integer::parseInt)
				.collect(Collectors.toList());

			List<Integer> Bs = Arrays.stream(getLine(bufferedReader).split(" "))
				.map(Integer::parseInt)
				.collect(Collectors.toList());

			int total = 0;

			List<Integer> ones = new ArrayList<>();
			List<Integer> twos = new ArrayList<>();
			for (int n = 0; n < N; n++) {
				if (Bs.get(n) == 1) {
					ones.add(As.get(n));
				} else {
					twos.add(As.get(n));
				}
			}

			Collections.sort(ones);
			Collections.sort(twos);
			int i1 = ones.size() - 1, i2 = twos.size() - 1;
			long onesSum = 0, twosSum = 0;

			while (M > 0) {
				boolean choose1 = false, choose2 = false;
				if (i1 < 0 && i2 < 0) {
					total = -1;
					break;
				} else if (i1 < 0) {
					choose2 = true;
				} else if (i2 < 0) {
					choose1 = true;
				} else {
					if (M - ones.get(i1) <= 0) {
						choose1 = true;
					} else if (M - twos.get(i2) <= 0) {
						choose2 = true;
					} else if (ones.get(i1) >= twos.get(i2)) {
						choose1 = true;
					} else if (i1 == 0) {
						choose2 = true;
					} else if (ones.get(i1 - 1) + ones.get(i1) < twos.get(i2)) {
						choose2 = true;
					} else {
						choose1 = true;
					}
				}

				if (choose1) {
					M -= ones.get(i1);
					total += 1;
					i1--;
				}
				if (choose2) {
					M -= twos.get(i2);
					total += 2;
					i2--;
				}
				

			}

			System.out.println(total);
		}

	}
}