Ivan has got an array of n non-negative integers a1, a2, ..., an. Ivan knows that the array is sorted in the non-decreasing order. 

Ivan wrote out integers 2a1, 2a2, ..., 2an on a piece of paper. Now he wonders, what minimum number of integers of form 2b (b ≥ 0) need to be added to the piece of paper so that the sum of all integers written on the paper equalled 2v - 1 for some integer v (v ≥ 0). 

Help Ivan, find the required quantity of numbers.

Input

The first line contains integer n (1 ≤ n ≤ 105). The second input line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 2·109). It is guaranteed that a1 ≤ a2 ≤ ... ≤ an.

Output

Print a single integer — the answer to the problem.

Examples

Input

4
0 1 1 1


Output

0


Input

1
3


Output

3

Note

In the first sample you do not need to add anything, the sum of numbers already equals 23 - 1 = 7.

In the second sample you need to add numbers 20, 21, 22.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Main {
	
	public static void main(String[] args) {
		try {
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			PrintWriter pw = new PrintWriter(System.out);

			int n = Integer.parseInt(br.readLine());
			int[] a = new int[n+1];
			a[n] = Integer.MAX_VALUE;
			String[] param = br.readLine().split(" ");
			for (int i = 0; i < n; i++) a[i] = Integer.parseInt(param[i]);
			
			int[] b = new int[n];
			int count = 0;
			int continuous = 0;
			int current = -1;
			for (int i = 0; i <= n; i++) {
				if (current < a[i]) {
					if (continuous == 0) {
						current = a[i];
						continuous++;
					} else {
						if (continuous % 2 > 0) b[count++] = current;
						continuous /= 2;
						current++;
						i--;
					}
				} else {
					continuous++;
				}
			}

			int answer = 0;
			int now = 0;
			for (int i = 0; i < count; i++) {
				answer += b[i]-now;
				now = b[i]+1;
			}

			pw.println(answer);
			
			pw.close();
			br.close();
		} catch (IOException e) {
			e.printStackTrace();
			return;
		}
	}
}
