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.awt.Point;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Stack;
import java.util.StringTokenizer;
import static java.lang.Math.*;
@SuppressWarnings("unused")

public class round184C {
    static class pair implements Comparable<pair>{
        int pow, count;
        public pair(int p, int c){
            pow = p;
            count = c;
        }
        public int compareTo(pair p){
            if(this.pow < p.pow)
                return -1;
            if(this.pow > p.pow)
                return 1;
            return 0;
        }
    }
    static BufferedReader br = new BufferedReader(new InputStreamReader(
            System.in));
    static StringTokenizer st = new StringTokenizer("");

    static int nextInt() throws Exception {
        return Integer.parseInt(next());
    }

    static String next() throws Exception {
        while (true) {
            if (st.hasMoreTokens()) {
                return st.nextToken();
            }
            String s = br.readLine();
            if (s == null) {
                return null;
            }
            st = new StringTokenizer(s);
        }
    }
    public static void main(String[] args)throws Exception {
        int n = nextInt();
        int [] powers = new int [n];
        for(int i = 0 ; i < n ; ++i)
            powers[i] = nextInt();
        Stack<Point> s = new Stack<Point>();
        for(int i = 0 ; i < n ;){
            int j;
            int cnt = 1;
            for(j = i + 1 ; j < n && powers[i] == powers[j] ; ++j, ++cnt);          
            s.push(new Point(powers[i], cnt));
            i = j;
        }
        Stack<Point> work = new Stack<Point>();     
        while(!s.isEmpty())
            work.push(s.pop());     
        HashSet<Integer> done = new HashSet<Integer>();
        while(!work.isEmpty()){
            Point p = work.pop();           
            if(work.isEmpty() && p.y == 1){
                done.add(p.x);
                break;
            }
            if(p.y % 2 == 1)
                done.add(p.x);                          
            if(!work.isEmpty() && work.peek().x == p.x + 1){                
                int last = work.pop().y;                
                work.push(new Point(p.x + 1, last + (p.y / 2)));
            }else{
                int nxt_count = p.y / 2;
                int nxt = p.x + 1;
                while(nxt_count > 0){
                    if(nxt_count % 2 == 1)
                        done.add(nxt);
                    nxt_count /= 2;
                    nxt++;
                    if(!work.isEmpty() && work.peek().x == nxt){
                        int last = work.pop().y;
                        work.push(new Point(nxt, last + nxt_count));
                        break;
                    }
                }
            }
        }
        int mx = -1;
        for(Object o : done){
            Integer oo = (Integer) o;
            if(oo > mx)
                mx = oo;
        }
        System.out.println(mx - done.size() + 1);
    }
}
