#include <bits/stdc++.h>
using namespace std;
long long _i_n() {
  long long x = 0;
  char y = getchar();
  while (y <= '9' & y >= '0') x *= 10, x += y - '0', y = getchar();
  return x;
}
set<long long> s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  cout << setprecision(12);
  long long n = _i_n();
  while (n--) {
    long long a;
    for (a = _i_n(); s.find(a) != s.end(); a++) s.erase(a);
    s.insert(a);
  }
  cout << *s.rbegin() - s.size() + 1 << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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;
long long _i_n() {
  long long x = 0;
  char y = getchar();
  while (y <= '9' & y >= '0') x *= 10, x += y - '0', y = getchar();
  return x;
}
set<long long> s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  cout << setprecision(12);
  long long n = _i_n();
  while (n--) {
    long long a;
    for (a = _i_n(); s.find(a) != s.end(); a++) s.erase(a);
    s.insert(a);
  }
  cout << *s.rbegin() - s.size() + 1 << '\n';
  return 0;
}

```