#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 10;
long long n;
vector<long long> v;
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long c = 0;
  long long p = 0;
  cin >> n;
  long long a;
  cin >> a;
  v.push_back(a);
  long long cnt = 1;
  for (int i = 1; i < n; i++) {
    cin >> a;
    v.push_back(a);
    if (v[i] == v[i - 1])
      cnt++;
    else {
      cnt += p;
      long long k = v[i] - v[i - 1];
      for (int j = 0; j < k; j++) {
        if (cnt % 2 == 1) c++;
        cnt /= 2;
        if (!cnt) break;
      }
      p = cnt;
      cnt = 1;
    }
  }
  int num = v[n - 1];
  cnt += p;
  while (cnt != 0) {
    if (cnt % 2 == 1) c++;
    cnt /= 2;
    num++;
  }
  cout << num - c;
  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 int N = 5e3 + 10;
long long n;
vector<long long> v;
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long c = 0;
  long long p = 0;
  cin >> n;
  long long a;
  cin >> a;
  v.push_back(a);
  long long cnt = 1;
  for (int i = 1; i < n; i++) {
    cin >> a;
    v.push_back(a);
    if (v[i] == v[i - 1])
      cnt++;
    else {
      cnt += p;
      long long k = v[i] - v[i - 1];
      for (int j = 0; j < k; j++) {
        if (cnt % 2 == 1) c++;
        cnt /= 2;
        if (!cnt) break;
      }
      p = cnt;
      cnt = 1;
    }
  }
  int num = v[n - 1];
  cnt += p;
  while (cnt != 0) {
    if (cnt % 2 == 1) c++;
    cnt /= 2;
    num++;
  }
  cout << num - c;
  return 0;
}

```