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 D {
    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        int cases = scn.nextInt();
        while (cases-- > 0) {
            int numApps = scn.nextInt();
            int memory = scn.nextInt();
            int[] memories = new int[numApps];
            int[] convs = new int[numApps];
            List<Integer> ones = new ArrayList<>();
            List<Integer> twos = new ArrayList<>();
            int sum = 0;
            for (int i = 0; i < numApps; i++) {
                memories[i] = scn.nextInt();
                sum += memories[i];
            }
            for (int i = 0; i < numApps; i++) {
                convs[i] = scn.nextInt();
            }

            for (int i = 0; i < numApps; i++) {
                if (convs[i] == 1) ones.add(memories[i]);
                else twos.add(memories[i]);
            }
            Collections.sort(ones, Collections.reverseOrder());
            Collections.sort(twos, Collections.reverseOrder());
            int[] prefixOnes = new int[ones.size()];
            int[] prefixTwos = new int[twos.size()];
            for (int i = 0; i < ones.size(); i++) {
                if (i == 0) prefixOnes[i] = ones.get(0);
                else prefixOnes[i] = ones.get(i) + prefixOnes[i - 1];
            }

            for (int i = 0; i < twos.size(); i++) {
                if (i == 0) prefixTwos[i] = twos.get(0);
                else prefixTwos[i] = twos.get(i) + prefixTwos[i - 1];
            }
            int minConv = (int) 1e12;
            if (ones.isEmpty()) {
                int index = Arrays.binarySearch(prefixTwos, memory);
                if (index < 0) index = -index - 1;
                minConv = index * 2 + 2;

            }
            for (int i = 0; i < ones.size(); i++) {
                int memoryCleared = prefixOnes[i];
                int toBeCleared = memory - memoryCleared;
                int index = -1;
                if (toBeCleared > 0) {
                    index = Arrays.binarySearch(prefixTwos, toBeCleared);
                    if (index < 0) index = -index - 1;
                }
                int conv = i * 1 + 1 + index * 2 + 2;
                minConv = Integer.min(conv, minConv);
            }
            if (sum < memory) minConv = -1;
            System.out.println(minConv);
        }
    }


}
