#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
const long long int inf = 2e9 + 5;
double PI = 3.14159265358979323846;
void solve() {
  int n;
  cin >> n;
  int a[n];
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  int now = -1;
  int ans = 0;
  for (auto &x : m) {
    if (x.second > 1) {
      int next = x.second / 2;
      x.second = x.second % 2;
      m[x.first + 1] += next;
    }
    ans += (x.first - now - 1);
    if (x.second % 2 == 0) ans++;
    now = x.first;
  }
  cout << ans << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  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;
const long long int mod = 1e9 + 7;
const long long int inf = 2e9 + 5;
double PI = 3.14159265358979323846;
void solve() {
  int n;
  cin >> n;
  int a[n];
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  int now = -1;
  int ans = 0;
  for (auto &x : m) {
    if (x.second > 1) {
      int next = x.second / 2;
      x.second = x.second % 2;
      m[x.first + 1] += next;
    }
    ans += (x.first - now - 1);
    if (x.second % 2 == 0) ans++;
    now = x.first;
  }
  cout << ans << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```