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.
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int M = 100010;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const double Pi = 2 * acos(0.0);
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
inline int bit(int x) { return 1 << x; }
inline double dabs(double x) { return x > 0 ? x : (-x); }
inline int abs(int x) { return x > 0 ? x : (-x); }
inline int lowbit(int x) { return x & (-x); }
int n, t;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  int k = 0, cnt = 0, tot = 0;
  for (int i = 0; i < n; i++) {
    cin >> t;
    if (t == k)
      cnt++;
    else {
      while (t != k && cnt) {
        if (cnt & 1)
          cnt--;
        else
          tot++;
        cnt /= 2;
        k++;
      }
      if (k == t)
        cnt++;
      else {
        tot += t - k;
        k = t;
        cnt = 1;
      }
    }
  }
  while (cnt) {
    if (cnt & 1)
      cnt--;
    else
      tot++;
    cnt /= 2;
  }
  cout << tot << endl;
  return 0;
}
