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.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Contest690 {

    public static final int MOD = 1000000007;
    public static void main(String[] args)throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int t = Integer.parseInt(br.readLine());

        while(t-->0) {
            String s = br.readLine();
            String[] str = s.split(" ");

            int n = Integer.parseInt(str[0]);
            int m = Integer.parseInt(str[1]);

            ArrayList<Long> memory1 = new ArrayList<>();
            ArrayList<Long> memory2 = new ArrayList<>();
            int[] memory = new int[n];
            int[] points = new int[n];

            s = br.readLine();
            str = s.split(" ");
            for (int i = 0; i < n; i++) {
                memory[i] = Integer.parseInt(str[i]);
            }

            s = br.readLine();
            str = s.split(" ");
            for (int i = 0; i < n; i++) {
                points[i] = Integer.parseInt(str[i]);
            }

            for (int i = 0; i < n; i++) {
                if(points[i]==1) {
                    memory1.add((long)memory[i]);
                }
                else {
                    memory2.add((long)memory[i]);
                }
            }
            Collections.sort(memory1,Collections.reverseOrder());
            Collections.sort(memory2,Collections.reverseOrder());

            long sum = 0;
            for (int i = 0; i < memory1.size(); i++) {
                sum+=memory1.get(i);
                memory1.set(i,sum);
            }

            int pos = Integer.MAX_VALUE;
            if(sum>=m) {
                pos = Collections.binarySearch(memory1, (long)m);
                if (pos < 0) {
                    pos = Math.abs(pos)-1;
                }
                pos++;
            }

            long sum1 = 0;
            for (int i = 0; i < memory2.size(); i++) {
                sum1+=memory2.get(i);
                if(sum1>=m){
                    pos = Math.min(pos,(i+1)*2);
                    break;
                }
                int pos1 = Collections.binarySearch(memory1, m - sum1);
                if (pos1 < 0) {
                    pos1 = Math.abs(pos1) - 1;
                }
                if (pos1 == memory1.size()) {
                    continue;
                }
                pos1++;
                pos = Math.min(pos, (i + 1) * 2 + pos1);
            }

            if(sum+sum1<m){
                System.out.println(-1);
                continue;
            }

            System.out.println(pos);

        }

    }

}