#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
  cin >> n;
  int t = 0;
  set<long long> s;
  long long temp;
  set<long long>::iterator it;
  for (int i = 0; i < n; i++) {
    cin >> temp;
  aa:
    it = s.find(temp);
    if (it == s.end()) {
      s.insert(temp);
    } else {
      s.erase(it);
      temp++;
      goto aa;
    }
  }
  int sz = s.size();
  long long mm = 0;
  for (it = s.begin(); it != s.end(); it++) mm = max(mm, *it);
  cout << mm + 1 - sz << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
int main() {
  cin >> n;
  int t = 0;
  set<long long> s;
  long long temp;
  set<long long>::iterator it;
  for (int i = 0; i < n; i++) {
    cin >> temp;
  aa:
    it = s.find(temp);
    if (it == s.end()) {
      s.insert(temp);
    } else {
      s.erase(it);
      temp++;
      goto aa;
    }
  }
  int sz = s.size();
  long long mm = 0;
  for (it = s.begin(); it != s.end(); it++) mm = max(mm, *it);
  cout << mm + 1 - sz << endl;
  return 0;
}

```