#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  vector<int> v;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    long long int x;
    cin >> x;
    v.push_back(x);
  }
  bool cambio = true;
  while (cambio) {
    cambio = false;
    vector<int> w;
    for (int i = 0; i < (v.size()); i++) {
      int c = v[i];
      int r = 0;
      while (i < v.size() and v[i] == c) {
        i++;
        r++;
      }
      int a = 0;
      if (r > 1) cambio = true;
      while (r) {
        if (r % 2) w.push_back(c + a);
        r /= 2;
        a++;
      }
      i--;
    }
    v = w;
    sort(v.begin(), v.end());
  }
  int c = v[0], res = v[0];
  for (int i = 0; i < (v.size() - 1); i++) {
    res += v[i + 1] - c - 1;
    c = v[i + 1];
  }
  cout << res << endl;
}

### 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;
int main() {
  int n;
  vector<int> v;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    long long int x;
    cin >> x;
    v.push_back(x);
  }
  bool cambio = true;
  while (cambio) {
    cambio = false;
    vector<int> w;
    for (int i = 0; i < (v.size()); i++) {
      int c = v[i];
      int r = 0;
      while (i < v.size() and v[i] == c) {
        i++;
        r++;
      }
      int a = 0;
      if (r > 1) cambio = true;
      while (r) {
        if (r % 2) w.push_back(c + a);
        r /= 2;
        a++;
      }
      i--;
    }
    v = w;
    sort(v.begin(), v.end());
  }
  int c = v[0], res = v[0];
  for (int i = 0; i < (v.size() - 1); i++) {
    res += v[i + 1] - c - 1;
    c = v[i + 1];
  }
  cout << res << endl;
}

```