#include <bits/stdc++.h>
using namespace std;
map<long long, long long> ps;
void addone(long long tmp) {
  ps[tmp]++;
  if (ps[tmp] == 2) {
    ps[tmp] = 0;
    addone(tmp + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long tmp;
    cin >> tmp;
    addone(tmp);
  }
  long long mx = 0, cnt = 0;
  for (map<long long, long long>::iterator it = ps.begin(); it != ps.end();
       it++) {
    if (it->second) {
      mx = max(it->first, mx);
      cnt++;
    }
  }
  cout << (mx + 1) - cnt << endl;
}

### 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;
map<long long, long long> ps;
void addone(long long tmp) {
  ps[tmp]++;
  if (ps[tmp] == 2) {
    ps[tmp] = 0;
    addone(tmp + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long tmp;
    cin >> tmp;
    addone(tmp);
  }
  long long mx = 0, cnt = 0;
  for (map<long long, long long>::iterator it = ps.begin(); it != ps.end();
       it++) {
    if (it->second) {
      mx = max(it->first, mx);
      cnt++;
    }
  }
  cout << (mx + 1) - cnt << endl;
}

```