#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, dem, prev, ans, n, j, u;
  scanf("%d\n", &n);
  dem = 0;
  prev = 0;
  ans = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d", &u);
    if ((i == 1) || (u == prev))
      dem++;
    else {
      for (j = prev; j <= u - 1; j++) {
        if ((dem % 2) == 1) ans++;
        dem /= 2;
        if (dem == 0) break;
      }
      dem += 1;
    }
    prev = u;
  }
  while (dem != 0) {
    if ((dem % 2) == 1) ans++;
    dem /= 2;
    if (dem != 0) prev++;
  }
  cout << (prev + 1 - ans);
}

### Prompt

Your task is to create 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 i, dem, prev, ans, n, j, u;
  scanf("%d\n", &n);
  dem = 0;
  prev = 0;
  ans = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d", &u);
    if ((i == 1) || (u == prev))
      dem++;
    else {
      for (j = prev; j <= u - 1; j++) {
        if ((dem % 2) == 1) ans++;
        dem /= 2;
        if (dem == 0) break;
      }
      dem += 1;
    }
    prev = u;
  }
  while (dem != 0) {
    if ((dem % 2) == 1) ans++;
    dem /= 2;
    if (dem != 0) prev++;
  }
  cout << (prev + 1 - ans);
}

```