#include <bits/stdc++.h>
using namespace std;
int n, temp, ans;
vector<int> a;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    a.push_back(temp);
    if (a[a.size() - 2] == a[a.size() - 1]) {
      a[a.size() - 2]++;
      a.pop_back();
    }
  }
  bool edit = true;
  while (edit) {
    edit = false;
    for (int i = 0; i < a.size() - 1; i++) {
      if (a[i] == a[i + 1]) {
        a[i]++;
        a.erase(a.begin() + i + 1);
        edit = true;
      }
    }
    sort(a.begin(), a.end());
  }
  for (int i = 0; i < a.size() - 1; i++) {
    if (a[i] + 1 != a[i + 1]) {
      ans += a[i + 1] - a[i] - 1;
    }
  }
  ans += a[0];
  cout << ans;
  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 namespace std;
int n, temp, ans;
vector<int> a;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    a.push_back(temp);
    if (a[a.size() - 2] == a[a.size() - 1]) {
      a[a.size() - 2]++;
      a.pop_back();
    }
  }
  bool edit = true;
  while (edit) {
    edit = false;
    for (int i = 0; i < a.size() - 1; i++) {
      if (a[i] == a[i + 1]) {
        a[i]++;
        a.erase(a.begin() + i + 1);
        edit = true;
      }
    }
    sort(a.begin(), a.end());
  }
  for (int i = 0; i < a.size() - 1; i++) {
    if (a[i] + 1 != a[i + 1]) {
      ans += a[i + 1] - a[i] - 1;
    }
  }
  ans += a[0];
  cout << ans;
  return 0;
}

```