#include <bits/stdc++.h>
using ll = int64_t;
using namespace std;
int main() {
  ll n, a;
  cin >> n;
  ll lastA = -1;
  ll answer = 0;
  ll reserve = 0;
  ll current = 0;
  for (ll i = 0; i < n; i++) {
    cin >> a;
    if (lastA != a) {
      if (current == 0 && lastA != -1) answer++;
      while (reserve > 0 && a > lastA + 1) {
        if (reserve % 2 == 0) {
          answer++;
        } else {
          reserve--;
        }
        reserve /= 2;
        lastA++;
      }
      answer += (a - lastA - 1);
      current = reserve;
      lastA = a;
    }
    current++;
    if (current >= 2) {
      reserve += current / 2;
      current %= 2;
    }
  }
  while (reserve != 0) {
    if (current == 0) answer++;
    current = reserve;
    reserve = 0;
    if (current >= 2) {
      reserve += current / 2;
      current %= 2;
    }
  }
  cout << answer;
  return 0;
}

### Prompt

Develop a solution in cpp 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 ll = int64_t;
using namespace std;
int main() {
  ll n, a;
  cin >> n;
  ll lastA = -1;
  ll answer = 0;
  ll reserve = 0;
  ll current = 0;
  for (ll i = 0; i < n; i++) {
    cin >> a;
    if (lastA != a) {
      if (current == 0 && lastA != -1) answer++;
      while (reserve > 0 && a > lastA + 1) {
        if (reserve % 2 == 0) {
          answer++;
        } else {
          reserve--;
        }
        reserve /= 2;
        lastA++;
      }
      answer += (a - lastA - 1);
      current = reserve;
      lastA = a;
    }
    current++;
    if (current >= 2) {
      reserve += current / 2;
      current %= 2;
    }
  }
  while (reserve != 0) {
    if (current == 0) answer++;
    current = reserve;
    reserve = 0;
    if (current >= 2) {
      reserve += current / 2;
      current %= 2;
    }
  }
  cout << answer;
  return 0;
}

```