#include <bits/stdc++.h>
const int inf = 1000005;
using namespace std;
map<long long, int> mp;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    if (!((mp).find(x) != (mp).end()))
      mp[x] = 1;
    else
      mp[x]++;
  }
  long long ans = 0LL;
  for (auto &y : mp) {
    int x = y.second;
    y.second = 0;
    while (x) {
      int indx = (x & (-x));
      x -= indx;
      indx = log2(indx);
      if (!((mp).find(y.first + indx) != (mp).end()))
        mp[y.first + indx] = 1;
      else
        mp[y.first + indx]++;
    }
    ans += y.second ? 1LL : 0LL;
  }
  cout << ((mp.rbegin()->first) + 1 - ans) << "\n";
  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>
const int inf = 1000005;
using namespace std;
map<long long, int> mp;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    if (!((mp).find(x) != (mp).end()))
      mp[x] = 1;
    else
      mp[x]++;
  }
  long long ans = 0LL;
  for (auto &y : mp) {
    int x = y.second;
    y.second = 0;
    while (x) {
      int indx = (x & (-x));
      x -= indx;
      indx = log2(indx);
      if (!((mp).find(y.first + indx) != (mp).end()))
        mp[y.first + indx] = 1;
      else
        mp[y.first + indx]++;
    }
    ans += y.second ? 1LL : 0LL;
  }
  cout << ((mp.rbegin()->first) + 1 - ans) << "\n";
  return 0;
}

```