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.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.StringTokenizer;

public class CleaningThePhone implements Closeable {

  private InputReader in = new InputReader(System.in);
  private PrintWriter out = new PrintWriter(System.out);

  public void solve() {
    int t = in.ni();
    while (t-- > 0) {
      int N = in.ni(), M = in.ni();
      List<Long> x = new ArrayList<>(), y = new ArrayList<>();
      long[] memory = new long[N];
      for (int i = 0; i < N; i++) {
        memory[i] = in.nl();
      }
      for (int i = 0; i < N; i++) {
        int type = in.ni();
        if (type == 1) {
          x.add(memory[i]);
        } else {
          y.add(memory[i]);
        }
      }
      x.sort(Comparator.reverseOrder());
      y.sort(Comparator.reverseOrder());
      int n = x.size();
      long[] one = new long[n];
      for (int i = 0; i < n; i++) {
        one[i] = x.get(i);
        if (i > 0) {
          one[i] += one[i - 1];
        }
      }
      int m = y.size();
      long[] two = new long[m];
      for (int i = 0; i < m; i++) {
        two[i] = y.get(i);
        if (i > 0) {
          two[i] += two[i - 1];
        }
      }

      int result = Integer.MAX_VALUE;
      long need = M;
      if (m >= 1 && two[m - 1] >= need) {
        int left = 0, right = m - 1, idx = m;
        while (left <= right) {
          int mid = left + (right - left) / 2;
          if (two[mid] >= need) {
            idx = Math.min(idx, mid);
            right = mid - 1;
          } else {
            left = mid + 1;
          }
        }
        result = Math.min(result, 2 * (idx + 1));
      }

      for (int i = 0; i < n; i++) {
        need = M - one[i];
        if (need <= 0) {
          result = Math.min(result, i + 1);
        } else {
          if (m >= 1 && two[m - 1] >= need) {
            int left = 0, right = m - 1, idx = m;
            while (left <= right) {
              int mid = left + (right - left) / 2;
              if (two[mid] >= need) {
                idx = Math.min(idx, mid);
                right = mid - 1;
              } else {
                left = mid + 1;
              }
            }
            result = Math.min(result, (i + 1) + 2 * (idx + 1));
          }
        }
      }
      out.println(result == Integer.MAX_VALUE ? -1 : result);
    }
  }

  @Override
  public void close() throws IOException {
    in.close();
    out.close();
  }

  static class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
      reader = new BufferedReader(new InputStreamReader(stream), 32768);
      tokenizer = null;
    }

    public String next() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public int ni() {
      return Integer.parseInt(next());
    }

    public long nl() {
      return Long.parseLong(next());
    }

    public void close() throws IOException {
      reader.close();
    }
  }

  public static void main(String[] args) throws IOException {
    try (CleaningThePhone instance = new CleaningThePhone()) {
      instance.solve();
    }
  }
}
