#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int sum = a[0], count = 0, maxi = a[0];
  for (int i = 1; i < n; i++) {
    if (a[i] == a[i - 1])
      count++;
    else {
      if (count) {
        if ((count + 1) & 1)
          ;
        else
          sum++;
        int t = a[i] - a[i - 1] - 1;
        count++;
        count /= 2;
        while (t-- && count) {
          if (count % 2 == 0) sum++;
          count /= 2;
        }
        if (t != -1) sum += t + 1;
        maxi = a[i];
      } else {
        sum += a[i] - a[i - 1] - 1;
      }
    }
  }
  if (count) {
    count++;
    if (count % 2 == 0) sum++;
    while (count /= 2) {
      if (count % 2 == 0) sum++;
    }
  }
  cout << sum;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int sum = a[0], count = 0, maxi = a[0];
  for (int i = 1; i < n; i++) {
    if (a[i] == a[i - 1])
      count++;
    else {
      if (count) {
        if ((count + 1) & 1)
          ;
        else
          sum++;
        int t = a[i] - a[i - 1] - 1;
        count++;
        count /= 2;
        while (t-- && count) {
          if (count % 2 == 0) sum++;
          count /= 2;
        }
        if (t != -1) sum += t + 1;
        maxi = a[i];
      } else {
        sum += a[i] - a[i - 1] - 1;
      }
    }
  }
  if (count) {
    count++;
    if (count % 2 == 0) sum++;
    while (count /= 2) {
      if (count % 2 == 0) sum++;
    }
  }
  cout << sum;
  return 0;
}

```