#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 22;
long long n, res, tmp, cnt, mn;
set<pair<int, int> > s;
set<pair<int, int> >::iterator it;
map<int, bool> mark;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 1, x; i <= n; i++) {
    cin >> x;
    if (tmp == x)
      cnt++;
    else {
      if (tmp != -1) s.insert(make_pair(tmp, cnt));
      tmp = x;
      cnt = 1;
    }
  }
  if (cnt) s.insert(make_pair(tmp, cnt));
  for (it = s.begin(); it != s.end(); it++) {
    long long x = (*it).first, ted = (*it).second;
    if (!mark[x]) {
      res += x - mn;
      mn = x + 1;
    }
    if (ted % 2 == 0 or (mark[x] and ted % 2 == 1)) res++;
    mark[x] = true;
    if (ted > 1) {
      s.insert(make_pair(x + 1, ted / 2));
    }
  }
  cout << res;
  return 0;
}

### Prompt

Generate 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;
const int MAXN = 1e5 + 22;
long long n, res, tmp, cnt, mn;
set<pair<int, int> > s;
set<pair<int, int> >::iterator it;
map<int, bool> mark;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 1, x; i <= n; i++) {
    cin >> x;
    if (tmp == x)
      cnt++;
    else {
      if (tmp != -1) s.insert(make_pair(tmp, cnt));
      tmp = x;
      cnt = 1;
    }
  }
  if (cnt) s.insert(make_pair(tmp, cnt));
  for (it = s.begin(); it != s.end(); it++) {
    long long x = (*it).first, ted = (*it).second;
    if (!mark[x]) {
      res += x - mn;
      mn = x + 1;
    }
    if (ted % 2 == 0 or (mark[x] and ted % 2 == 1)) res++;
    mark[x] = true;
    if (ted > 1) {
      s.insert(make_pair(x + 1, ted / 2));
    }
  }
  cout << res;
  return 0;
}

```