Problem: Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it even affects him while solving problems. At the moment, Mike has two sequences of positive integers A = [a1, a2, ..., an] and B = [b1, b2, ..., bn] of length n each which he uses to ask people some quite peculiar questions.

To test you on how good are you at spotting inequality in life, he wants you to find an "unfair" subset of the original sequence. To be more precise, he wants you to select k numbers P = [p1, p2, ..., pk] such that 1 ≤ pi ≤ n for 1 ≤ i ≤ k and elements in P are distinct. Sequence P will represent indices of elements that you'll select from both sequences. He calls such a subset P "unfair" if and only if the following conditions are satisfied: 2·(ap1 + ... + apk) is greater than the sum of all elements from sequence A, and 2·(bp1 + ... + bpk) is greater than the sum of all elements from the sequence B. Also, k should be smaller or equal to <image> because it will be to easy to find sequence P if he allowed you to select too many elements!

Mike guarantees you that a solution will always exist given the conditions described above, so please help him satisfy his curiosity!

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of elements in the sequences. 

On the second line there are n space-separated integers a1, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

On the third line there are also n space-separated integers b1, ..., bn (1 ≤ bi ≤ 109) — elements of sequence B.

Output

On the first line output an integer k which represents the size of the found subset. k should be less or equal to <image>.

On the next line print k integers p1, p2, ..., pk (1 ≤ pi ≤ n) — the elements of sequence P. You can print the numbers in any order you want. Elements in sequence P should be distinct.

Example

Input

5
8 7 4 8 3
4 2 5 3 7


Output

3
1 4 5

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author toshif
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        MyReader in = new MyReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        public void solve(int testNumber, MyReader in, PrintWriter out) {
            int n = in.nextInt();
            long[] a = new long[n];
            long[] b = new long[n];
            long asum = 0, bsum = 0;
            for (int i = 0; i < n; i++) {
                a[i] = in.nextLong();
                asum += a[i];
            }
            for (int i = 0; i < n; i++) {
                b[i] = in.nextLong();
                bsum += b[i];
            }

            List<P> ps = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                P p = new P();
                p.a = a[i];
                p.b = b[i];
                p.idx = i + 1;
                ps.add(p);
            }
            Collections.sort(ps);
            Collections.reverse(ps);

            // n 6 , k 4
            // n 27, k 14
            // n 100, k 51
            int k = n / 2 + 1;
            List<P> ks = new ArrayList<>();

            ks.add(ps.remove(0));

            for (int i = 0; i < n / 2; i++) {
                if (i * 2 + 1 >= ps.size()) {
                    ks.add(ps.get(i * 2));
                    break;
                }

                P p1 = ps.get(i * 2);
                P p2 = ps.get(i * 2 + 1);

                if (p1.b > p2.b) {
                    ks.add(p1);
                } else {
                    ks.add(p2);
                }
            }

            long a2 = 0, b2 = 0;
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < k; i++) {
                sb.append(ks.get(i).idx);
                sb.append(" ");

                a2 += ks.get(i).a * 2;
                b2 += ks.get(i).b * 2;
            }

            System.err.printf("asum=%s, a2=%s, ok=%s \n", asum, a2, asum < a2);
            System.err.printf("bsum=%s, b2=%s, ok=%s \n", bsum, b2, bsum < b2);

            out.println(k);
            out.println(sb.toString());
        }

        class P implements Comparable<P> {
            long a;
            long b;
            int idx;


            public int compareTo(P o) {
                if (Long.compare(a, o.a) != 0) return Long.compare(a, o.a);
                if (Long.compare(b, o.b) != 0) return Long.compare(b, o.b);
                return Long.compare(idx, o.idx);
            }

        }

    }

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

        public MyReader(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());
        }

    }
}