#include <bits/stdc++.h>
using namespace std;
int main() {
  map<int, int> v;
  map<int, int>::iterator ite;
  pair<map<int, int>::iterator, bool> pai;
  int n;
  int i;
  int a, b;
  cin >> n;
  for (i = 0; i < n; i++) {
    scanf("%d", &a);
    pai = v.insert(pair<int, int>(a, 1));
    if (pai.second == false) {
      pai.first->second++;
    }
  }
  for (ite = v.begin(); ite != v.end(); ite++) {
    a = ite->second;
    ite->second = a & 1;
    i = 1;
    while (1) {
      b = 1 << i;
      if (b > a) break;
      if (b & a) {
        pai = v.insert(pair<int, int>(ite->first + i, 1));
        if (pai.second == false) {
          pai.first->second++;
        }
      }
      i++;
    }
  }
  a = 0;
  for (ite = v.begin(); ite != v.end(); ite++) {
    if (ite->second != 0) a++;
    b = ite->first;
  }
  cout << b + 1 - a << endl;
}

### Prompt

Please create a solution in cpp 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() {
  map<int, int> v;
  map<int, int>::iterator ite;
  pair<map<int, int>::iterator, bool> pai;
  int n;
  int i;
  int a, b;
  cin >> n;
  for (i = 0; i < n; i++) {
    scanf("%d", &a);
    pai = v.insert(pair<int, int>(a, 1));
    if (pai.second == false) {
      pai.first->second++;
    }
  }
  for (ite = v.begin(); ite != v.end(); ite++) {
    a = ite->second;
    ite->second = a & 1;
    i = 1;
    while (1) {
      b = 1 << i;
      if (b > a) break;
      if (b & a) {
        pai = v.insert(pair<int, int>(ite->first + i, 1));
        if (pai.second == false) {
          pai.first->second++;
        }
      }
      i++;
    }
  }
  a = 0;
  for (ite = v.begin(); ite != v.end(); ite++) {
    if (ite->second != 0) a++;
    b = ite->first;
  }
  cout << b + 1 - a << endl;
}

```