You are given a permutation p of length n. Remove one element from permutation to make the number of records the maximum possible.

We remind that in a sequence of numbers a1, a2, ..., ak the element ai is a record if for every integer j (1 ≤ j < i) the following holds: aj < ai. 

Input

The first line contains the only integer n (1 ≤ n ≤ 105) — the length of the permutation.

The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation. All the integers are distinct.

Output

Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.

Examples

Input

1
1


Output

1


Input

5
5 1 2 3 4


Output

5

Note

In the first example the only element can be removed.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.io.InputStream;

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

    static class TaskC {
        TreeSet<Integer> set;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int N = in.nextInt();
            int[] P = new int[N];
            for (int i = 0; i < N; i++) {
                P[i] = in.nextInt();
            }
            set = new TreeSet<>();
            set.add(Integer.MIN_VALUE);
            set.add(Integer.MIN_VALUE + 1);

            boolean[] record = new boolean[N + 1];
            int[] cnt = new int[N + 1];

            for (int i = 0; i < N; i++) {
                update(P, i - 1);

                if (largest() < P[i]) {
                    record[P[i]] = true;
                } else if (secondLargest() < P[i]) {
                    cnt[largest()]++;
                }
            }

            for (int i = 1; i <= N; i++) {
                if (record[i]) {
                    cnt[i]--;
                }
            }

            int best = 1;
            for (int i = 2; i <= N; i++) {
                if (cnt[i] > cnt[best]) {
                    best = i;
                }
            }
            out.println(best);
        }

        private int secondLargest() {
            return set.first();
        }

        private int largest() {
            return set.last();
        }

        private void update(int[] P, int idx) {
            if (idx < 0) return;
            set.add(P[idx]);
            if (set.size() > 2) set.pollFirst();
        }

    }

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

    }
}

