#include <bits/stdc++.h>
using namespace std;
map<long long int, long long int> mp;
vector<long long int> ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, i, val, freq, curr, k, p, answer = 0;
  pair<long long int, long long int> pp;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> k;
    mp[k]++;
  }
  for (auto it = mp.begin(); it != mp.end(); it++) {
    pp = (*it);
    val = pp.first;
    freq = pp.second;
    if (freq % 2 == 1) {
      ans.push_back(val);
    }
    freq /= 2;
    if (freq != 0) mp[val + 1] += freq;
  }
  p = -1;
  for (i = 0; i < ans.size(); i++) {
    curr = ans[i];
    answer = answer + (curr - p - 1);
    p = curr;
  }
  cout << answer;
  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;
map<long long int, long long int> mp;
vector<long long int> ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, i, val, freq, curr, k, p, answer = 0;
  pair<long long int, long long int> pp;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> k;
    mp[k]++;
  }
  for (auto it = mp.begin(); it != mp.end(); it++) {
    pp = (*it);
    val = pp.first;
    freq = pp.second;
    if (freq % 2 == 1) {
      ans.push_back(val);
    }
    freq /= 2;
    if (freq != 0) mp[val + 1] += freq;
  }
  p = -1;
  for (i = 0; i < ans.size(); i++) {
    curr = ans[i];
    answer = answer + (curr - p - 1);
    p = curr;
  }
  cout << answer;
  return 0;
}

```