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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;

/*
@author kalZor
 */
public class TaskD {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solver solver = new Solver();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class Solver {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            ArrayList<Integer> a = new ArrayList<>();
            ArrayList<Integer> b = new ArrayList<>();
            int[] ta = in.nextArray(n);
            int[] tb = in.nextArray(n);
            long totSum = 0;
            for(int i=0;i<n;i++){
                if(tb[i]==1) {
                    a.add(ta[i]);
                    totSum+=ta[i];
                }
                else{
                    b.add(ta[i]);
                    totSum+=ta[i];
                }
            }
            if(totSum<m){
                out.println(-1);
                return;
            }
            Collections.sort(a, Collections.reverseOrder());
            Collections.sort(b, Collections.reverseOrder());
//            out.println(a);

            long[] prefa = new long[a.size()];
            long[] prefb = new long[b.size()];
            for(int i=0;i<prefa.length;i++){
                if(i==0) prefa[i] = a.get(i);
                else prefa[i] = prefa[i-1]+a.get(i);
            }
            for(int i=0;i<prefb.length;i++){
                if(i==0) prefb[i] = b.get(i);
                else prefb[i] = prefb[i-1]+b.get(i);
            }
            int ans = Integer.MAX_VALUE;
            if(prefb.length==0||prefa.length==0){
                if(prefb.length==0){
                    for(int i=0;i<prefa.length;i++){
                        if(prefa[i]>=m){
                            out.println(i+1);
                            return;
                        }
                    }
                }
                if(prefa.length==0){
                    for(int i=0;i<prefb.length;i++){
                        if(prefb[i]>=m){
                            out.println((i+1)*2);
                            return;
                        }
                    }
                }
            }
            for(int i=0;i<prefa.length;i++){
                long sum = prefa[i];
                if(prefb.length>0&&m-prefa[i]>prefb[prefb.length-1]) continue;
                if(m-prefa[i]<=0) {
                    ans =Math.min(ans,i+1);
                    continue;
                }
                int index = Arrays.binarySearch(prefb,(m-prefa[i]));
                if(index<0){
                    index = -(index+1);
//                    index++;
                }
                ans = Math.min(ans,(i+1)+(2*(index+1)));
            }
            out.println(ans);

        }
    }

    class Pair implements Comparable<Pair>{
        int a;
        int b;
        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
        public int compareTo(Pair o) {
            if(this.a==o.a)
                return this.b - o.b;
            return this.a - o.a;
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader(InputStream stream){
            br = new BufferedReader(new
                InputStreamReader(stream));
        }

        public String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        public double nextDouble(){
            return Double.parseDouble(next());
        }

        public String nextLine(){
            String str = "";
            try{
                str = br.readLine();
            }
            catch (IOException e){
                e.printStackTrace();
            }
            return str;
        }

        public int[] nextArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
                return a;
        }
    }
}

