#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  deque<int> d1, d2;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    d1.push_back(a);
  }
  int answer = 0;
  int current = 0, count = 0;
  while (1) {
    if (d1.empty() || current < d1.front()) {
      while (count && (d1.empty() || current < d1.front())) {
        if (count % 2 == 0) answer++;
        current++, count >>= 1;
      }
      if (!d1.empty()) {
        answer += d1.front() - current;
        current = d1.front();
      }
      if (d1.empty() && count == 0) break;
    } else {
      while (!d1.empty() && current == d1.front()) d1.pop_front(), count++;
    }
  }
  cout << answer;
  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 main() {
  int n;
  cin >> n;
  deque<int> d1, d2;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    d1.push_back(a);
  }
  int answer = 0;
  int current = 0, count = 0;
  while (1) {
    if (d1.empty() || current < d1.front()) {
      while (count && (d1.empty() || current < d1.front())) {
        if (count % 2 == 0) answer++;
        current++, count >>= 1;
      }
      if (!d1.empty()) {
        answer += d1.front() - current;
        current = d1.front();
      }
      if (d1.empty() && count == 0) break;
    } else {
      while (!d1.empty() && current == d1.front()) d1.pop_front(), count++;
    }
  }
  cout << answer;
  return 0;
}

```