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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class A {

	public static void main(String[] args) throws IOException {
		FastScanner sc = new FastScanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int[] a = new int[n];
		int[] blocker = new int[n];
		int[] cnt = new int[n + 1];
		for(int i=1;i<=n;i++) cnt[i] = 0;
		
		TreeSet<Integer> ts = new TreeSet<Integer>();
		
		for(int i=0;i<n;i++) a[i] = sc.nextInt();
		for(int i=0;i<n;i++) {
			blocker[i] = 0;
			Integer e = ts.higher(a[i]);
			if(e == null) cnt[a[i]] = -1;
			if(!ts.isEmpty() && e != null && e.equals(ts.last())) blocker[i] = e;
			ts.add(a[i]);
		}
		
		for(int i=0;i<n;i++) cnt[blocker[i]]++;
		
		int mx = -100, x = 1;
		for(int i=1;i<=n;i++) if(cnt[i] > mx) {
			mx = cnt[i]; x = i;
		}			
		
		pw.println(x);
		
		pw.close();
		sc.close();
	}

}

class FastScanner {
	BufferedReader br;
	StringTokenizer st;

	public FastScanner(InputStream is) {
		br = new BufferedReader(new InputStreamReader(is));
		st = null;
	}

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

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

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

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

	public void close() throws IOException {
		br.close();
	}
}