#include <bits/stdc++.h>
using namespace std;
int a[100002];
int main() {
  int n, i;
  int in;
  int sum;
  int innum;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  in = a[0];
  innum = 1;
  sum = 0;
  for (i = 1; i < n; i++) {
    if (a[i] == in)
      innum++;
    else {
      while (1) {
        if (innum % 2 == 1) sum++;
        innum /= 2;
        in++;
        if (innum == 0) {
          in = a[i];
          innum = 1;
          break;
        } else if (a[i] == in) {
          innum++;
          break;
        }
      }
    }
  }
  while (1) {
    if (innum % 2 == 1) sum++;
    innum /= 2;
    in++;
    if (innum == 0) {
      innum = 1;
      break;
    }
  }
  printf("%d\n", in - 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 a[100002];
int main() {
  int n, i;
  int in;
  int sum;
  int innum;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  in = a[0];
  innum = 1;
  sum = 0;
  for (i = 1; i < n; i++) {
    if (a[i] == in)
      innum++;
    else {
      while (1) {
        if (innum % 2 == 1) sum++;
        innum /= 2;
        in++;
        if (innum == 0) {
          in = a[i];
          innum = 1;
          break;
        } else if (a[i] == in) {
          innum++;
          break;
        }
      }
    }
  }
  while (1) {
    if (innum % 2 == 1) sum++;
    innum /= 2;
    in++;
    if (innum == 0) {
      innum = 1;
      break;
    }
  }
  printf("%d\n", in - sum);
  return 0;
}

```