#include <bits/stdc++.h>
using namespace std;
int main() {
  int k;
  scanf("%d", &k);
  int a_now = 0;
  int digit_now = 0;
  int required = 0;
  for (int i = 0; i < k; ++i) {
    int a;
    scanf("%d", &a);
    if (a > a_now) {
      for (int d = a_now; d < a; ++d) {
        if (digit_now % 2 == 0) {
          ++required;
        }
        digit_now /= 2;
        if (digit_now == 0) {
          required += a - d - 1;
          break;
        }
      }
      a_now = a;
    }
    ++digit_now;
  }
  for (; digit_now > 0; digit_now /= 2) {
    if (digit_now % 2 == 0) {
      ++required;
    }
  }
  printf("%d\n", required);
  return 0;
}

### Prompt

Develop a solution in Cpp 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() {
  int k;
  scanf("%d", &k);
  int a_now = 0;
  int digit_now = 0;
  int required = 0;
  for (int i = 0; i < k; ++i) {
    int a;
    scanf("%d", &a);
    if (a > a_now) {
      for (int d = a_now; d < a; ++d) {
        if (digit_now % 2 == 0) {
          ++required;
        }
        digit_now /= 2;
        if (digit_now == 0) {
          required += a - d - 1;
          break;
        }
      }
      a_now = a;
    }
    ++digit_now;
  }
  for (; digit_now > 0; digit_now /= 2) {
    if (digit_now % 2 == 0) {
      ++required;
    }
  }
  printf("%d\n", required);
  return 0;
}

```