#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  int a, t;
  int nb = 0, maxi = -1;
  priority_queue<int> feuille;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    while (a != -1) {
      while (not feuille.empty() and feuille.top() > -a) {
        nb++;
        maxi = max(maxi, -feuille.top());
        feuille.pop();
      }
      if (feuille.empty() or feuille.top() != -a) {
        feuille.push(-a);
        a = -1;
      } else {
        feuille.pop();
        a = a + 1;
      }
    }
  }
  while (not feuille.empty()) {
    nb++;
    maxi = max(maxi, -feuille.top());
    feuille.pop();
  }
  printf("%d\n", maxi - nb + 1);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  int a, t;
  int nb = 0, maxi = -1;
  priority_queue<int> feuille;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    while (a != -1) {
      while (not feuille.empty() and feuille.top() > -a) {
        nb++;
        maxi = max(maxi, -feuille.top());
        feuille.pop();
      }
      if (feuille.empty() or feuille.top() != -a) {
        feuille.push(-a);
        a = -1;
      } else {
        feuille.pop();
        a = a + 1;
      }
    }
  }
  while (not feuille.empty()) {
    nb++;
    maxi = max(maxi, -feuille.top());
    feuille.pop();
  }
  printf("%d\n", maxi - nb + 1);
  return 0;
}

```