#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, t, n, m, k, l, r, temp, mini, maxi = -1, flag, result, cnt;
  scanf("%d", &n);
  stack<int> S;
  S.push(-1);
  for (i = 0; i < n; i++) {
    scanf("%d", &k);
    S.push(k);
    maxi = max(maxi, k);
    temp = S.top();
    S.pop();
    while (S.top() == temp) {
      S.pop();
      S.push(temp + 1);
      maxi = max(maxi, temp + 1);
      temp = S.top();
      S.pop();
    }
    S.push(temp);
    maxi = max(maxi, temp);
  }
  printf("%d\n", maxi + 2 - (int)S.size());
  return 0;
}

### Prompt

Please create a solution in cpp 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, j, t, n, m, k, l, r, temp, mini, maxi = -1, flag, result, cnt;
  scanf("%d", &n);
  stack<int> S;
  S.push(-1);
  for (i = 0; i < n; i++) {
    scanf("%d", &k);
    S.push(k);
    maxi = max(maxi, k);
    temp = S.top();
    S.pop();
    while (S.top() == temp) {
      S.pop();
      S.push(temp + 1);
      maxi = max(maxi, temp + 1);
      temp = S.top();
      S.pop();
    }
    S.push(temp);
    maxi = max(maxi, temp);
  }
  printf("%d\n", maxi + 2 - (int)S.size());
  return 0;
}

```