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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;

public class Main {

	public static void main(String [] args ) {
		try{
			String str;			
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			BufferedOutputStream bos = new BufferedOutputStream(System.out);
			String eol = System.getProperty("line.separator");
			byte [] eolb = eol.getBytes();
			byte[] spaceb= " ".getBytes();
			str  = br.readLine();
			int n = Integer.parseInt(str);
			long  [] ar = new long[n];
			int j=0;
			int s=0;
			int k =0;
			str = br.readLine();
			int length = str.length();
			while(j<length) {
				while(j<length) {
					if(str.charAt(j) == ' ') {
						break;
					}else {
						j++;
					}
				}
				ar[k] = Long.parseLong(str.substring(s,j)) ;
				k++;
				j++;
				s=j;			
			}
			TreeSet<Long> tree = new TreeSet<Long>();
			int iter = 0;
			while(iter<n) {
				if(tree.contains(ar[iter])) {
					tree.remove(ar[iter]);
					ar[iter]++;
				} else {
					tree.add(ar[iter]);
					iter++;
				}
			}
			Long last = tree.last();
			int size = tree.size();
			long ans = last - size + 1;
			bos.write(new Long(ans).toString().getBytes());
			bos.write(eolb);
			bos.flush();
		}  catch(IOException ioe) {
			ioe.printStackTrace();
		}
	}
}
