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.*;
import java.io.*;
public class MainClass {
    public static void main(String[] args) {
        Reader in = new Reader(System.in);
        StringBuilder stringBuilder = new StringBuilder();
        int t = in.nextInt();
        while (t-- > 0)
        {
            int n = in.nextInt();
            long m = in.nextLong();
            long ss = 0L;
            long[] A = new long[n];
            int[] B = new int[n];
            for (int i=0;i<n;i++)
                ss += A[i] = in.nextLong();
            for (int i=0;i<n;i++)
                B[i] = in.nextInt();
            ArrayList<Long> ones = new ArrayList<>();
            ArrayList<Long> twos = new ArrayList<>();
            for (int i=0;i<n;i++)
            {
                if (B[i] == 1)
                    ones.add(A[i]);
                else
                    twos.add(A[i]);
            }
            Collections.sort(ones, Collections.reverseOrder());
            Collections.sort(twos, Collections.reverseOrder());
            ArrayList<Long> sum = new ArrayList<>();
            if (twos.size() > 0)
                sum.add(twos.get(0));
            for (int i=1;i<twos.size();i++)
            {
                sum.add(sum.get(sum.size() - 1) + twos.get(i));
            }
            int k = twos.size();
            if (ss < m)
                stringBuilder.append("-1");
            else
            {
                long ans = Long.MAX_VALUE;
                long mm = 0L;
                for (int i=0;i<ones.size();i++)
                {
                    mm += ones.get(i);
                    long cost = i + 1;
                    long left = m - mm;
                    if (left <= 0L)
                    {
                        ans = Math.min(ans, cost);
                        continue;
                    }
                    int index = Collections.binarySearch(sum, left);
                    if (index < 0)
                        index = -index - 1;
                    if (index == k)
                        continue;
                    cost += 2L * (index + 1);
                    ans = Math.min(ans, cost);
                }
                int index = Collections.binarySearch(sum, m);
                if (index < 0)
                    index = -index - 1;
                if (index < k)
                {
                    ans = Math.min(ans, 2 * index + 2 * 1);
                }
                stringBuilder.append(ans);
            }
            stringBuilder.append("\n");
        }
        System.out.println(stringBuilder);
    }
}
class Reader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public Reader(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 nextInt() {
        return Integer.parseInt(next());
    }

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

}