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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Collections;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Xinyi Tao
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DCleaningThePhone solver = new DCleaningThePhone();
        solver.solve(1, in, out);
        out.close();
    }

    static class DCleaningThePhone {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int t = in.nextInt();
            for (int i = 0; i < t; i++) {
                int n = in.nextInt();
                int m = in.nextInt();
                int[] memory = new int[n];
                int[] point = new int[n];
                List<Integer> m1 = new ArrayList<>();
                List<Integer> m2 = new ArrayList<>();
                long s = 0;
                for (int j = 0; j < n; j++) {
                    memory[j] = in.nextInt();
                }
                for (int j = 0; j < n; j++) {
                    if (in.nextInt() == 1) {
                        m1.add(memory[j]);
                    } else {
                        m2.add(memory[j]);
                    }
                }
                m1.sort(Collections.reverseOrder());
                m2.sort(Collections.reverseOrder());

                Arrays.sort(point);
                int left;
                int r = Integer.MAX_VALUE;
                for (left = 0; left < m1.size(); left++) {
                    s += m1.get(left);
                    if (s >= m) {
                        r = left + 1;
                        break;
                    }
                }
                if (left == m1.size()) left = m1.size() - 1;
                int right;
                for (right = 0; right < m2.size(); right++) {
                    s += m2.get(right);
                    while (left >= 0 && s - m1.get(left) >= m) {
                        s -= m1.get(left);
                        left--;
                    }
                    if (s >= m) r = Math.min(r, left + 1 + 2 * (right + 1));
                }
                out.println(r == Integer.MAX_VALUE ? -1 : r);
            }
        }

    }
}

