#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int _n(n), i(0); i < _n; i++) cin >> a[i];
  map<long long, int> cnt;
  for (int _n(n), i(0); i < _n; i++) cnt[a[i]]++;
  set<long long> num((a).begin(), (a).end());
  for (long long x = -1;;) {
    set<long long>::iterator it = num.upper_bound(x);
    if (it == num.end()) break;
    x = *it;
    int s = cnt[x];
    cnt[x] = 0;
    for (long long y = x; s > 0; ++y, s /= 2) {
      if (s % 2) {
        cnt[y] += 1;
        num.insert(y);
      }
    }
  }
  int s = 0;
  long long mx = -1;
  for (map<long long, int>::iterator it = cnt.begin(); it != cnt.end(); ++it) {
    if (it->second == 0) continue;
    ++s;
    mx = max(mx, it->first);
  }
  cout << mx + 1 - s << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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);
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int _n(n), i(0); i < _n; i++) cin >> a[i];
  map<long long, int> cnt;
  for (int _n(n), i(0); i < _n; i++) cnt[a[i]]++;
  set<long long> num((a).begin(), (a).end());
  for (long long x = -1;;) {
    set<long long>::iterator it = num.upper_bound(x);
    if (it == num.end()) break;
    x = *it;
    int s = cnt[x];
    cnt[x] = 0;
    for (long long y = x; s > 0; ++y, s /= 2) {
      if (s % 2) {
        cnt[y] += 1;
        num.insert(y);
      }
    }
  }
  int s = 0;
  long long mx = -1;
  for (map<long long, int>::iterator it = cnt.begin(); it != cnt.end(); ++it) {
    if (it->second == 0) continue;
    ++s;
    mx = max(mx, it->first);
  }
  cout << mx + 1 - s << endl;
  return 0;
}

```