#include <bits/stdc++.h>
using namespace std;
int n;
int a[100010];
int st[100010];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int t = 0;
  st[t++] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    st[t] = a[i];
    while (t > 0 && st[t] == st[t - 1]) {
      st[t - 1] = st[t - 1] + 1;
      t--;
    }
    t++;
  }
  cout << st[0] + 1 - t;
  return 0;
}

### Prompt

Please formulate 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 n;
int a[100010];
int st[100010];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int t = 0;
  st[t++] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    st[t] = a[i];
    while (t > 0 && st[t] == st[t - 1]) {
      st[t - 1] = st[t - 1] + 1;
      t--;
    }
    t++;
  }
  cout << st[0] + 1 - t;
  return 0;
}

```