#include <bits/stdc++.h>
using namespace std;
int i, n, a[100005], m, cnt = 0;
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  i = 1;
  while (i < n) {
    int pos = upper_bound(a + i + 1, a + n + 1, a[i]) - a, t = a[i], c, j, cc;
    c = pos - i >> 1;
    cc = c;
    j = pos;
    while (cc) {
      cc--;
      j--;
      a[j] = t + 1;
    }
    i = j;
    while (c) {
      c--;
      j--;
      a[j] = -1;
    }
  }
  for (i = 1; i <= n; i++)
    if (a[i] != -1) cnt++;
  cout << a[n] - cnt + 1;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 i, n, a[100005], m, cnt = 0;
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  i = 1;
  while (i < n) {
    int pos = upper_bound(a + i + 1, a + n + 1, a[i]) - a, t = a[i], c, j, cc;
    c = pos - i >> 1;
    cc = c;
    j = pos;
    while (cc) {
      cc--;
      j--;
      a[j] = t + 1;
    }
    i = j;
    while (c) {
      c--;
      j--;
      a[j] = -1;
    }
  }
  for (i = 1; i <= n; i++)
    if (a[i] != -1) cnt++;
  cout << a[n] - cnt + 1;
  return 0;
}

```