#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  map<int, int> cnt;
  for (int i = 0; i < (int)(n); i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int last = -1;
  while (last < prev(cnt.end())->first) {
    auto p = *(cnt.upper_bound(last));
    if (p.second > 1) {
      cnt[p.first + 1] += p.second / 2;
      cnt[p.first] %= 2;
    }
    last = p.first;
  }
  int cnto = 0;
  for (auto p : cnt)
    if (p.second == 1) cnto++;
  cout << last + 1 - cnto << endl;
  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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  map<int, int> cnt;
  for (int i = 0; i < (int)(n); i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  int last = -1;
  while (last < prev(cnt.end())->first) {
    auto p = *(cnt.upper_bound(last));
    if (p.second > 1) {
      cnt[p.first + 1] += p.second / 2;
      cnt[p.first] %= 2;
    }
    last = p.first;
  }
  int cnto = 0;
  for (auto p : cnt)
    if (p.second == 1) cnto++;
  cout << last + 1 - cnto << endl;
  return 0;
}

```