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.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        new Solution().solve();
    }
}

class Solution {
    public void solve() throws Exception {
        // FastReader fr = new FastReader("input.txt");
        FastReader fr = new FastReader();
        FastWriter fw = new FastWriter();
        int t = fr.nextInt();
        while(t-->0){
            long n = fr.nextInt();
            long m = fr.nextInt();
            long[] a = new long[(int) n];
            long[] b = new long[(int) n];
            var list1 = new ArrayList<Long>();
            var list2 = new ArrayList<Long>();

            for(int i=0; i<n; i++)a[i] = fr.nextInt();
            for(int i=0; i<n; i++) {
                b[i] = fr.nextInt();
                if(b[i]==1)list1.add(a[i]);
                else list2.add(a[i]);
            }

            list1.sort(Comparator
                               .comparingLong(Long::longValue)
                               .reversed());
            list2.sort(Comparator
                               .comparingLong(Long::longValue)
                               .reversed());

            long pref = 0;
            long pref2 = list2.stream().mapToLong(Long::longValue).sum();
            long ans = Integer.MAX_VALUE;
            int j = list2.size();
            for(int i=0; i<=list1.size(); i++){
                while(j>0 && pref + pref2 - list2.get(j-1)>= m){
                    j--;
                    pref2 -= list2.get(j);
                }
                if(pref + pref2>=m){
                    ans = Math.min(ans, i + 2*j);
                }
                if(i<list1.size())
                    pref+=list1.get(i);
            }
            if(ans == Integer.MAX_VALUE)ans = -1;
            fw.println(ans);
        }

        fw.flush();
    }
}

class FastReader {
    private final BufferedReader br;
    private StringTokenizer st;

    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public FastReader(String filePath) throws FileNotFoundException {
        br = new BufferedReader(new InputStreamReader(new FileInputStream(filePath)));
    }

    public String nextString() throws IOException {
        if (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextString());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextString());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextString());
    }
}

class FastWriter {
    private final BufferedWriter bw;

    public FastWriter() {
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }

    public void println(Object o) throws IOException {
        bw.write(o.toString());
        println();
    }

    public void println() throws IOException {
        bw.write('\n');
    }

    public void printSp(Object o) throws IOException {
        bw.write(o.toString());
        bw.write(' ');
    }

    public void flush() throws IOException {
        bw.flush();
    }
}
