#include <bits/stdc++.h>
int N, p[200000];
int s, ans;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", p + i);
  s = 0;
  ans = p[1];
  for (int i = 1, pos = 0; i <= N;) {
    pos = p[i];
    while (i < N && p[i + 1] == p[i]) i++, s++;
    i++;
    s++;
    if (i > N) break;
    while (s && pos < p[i]) {
      if ((s & 1) == 0) ans++;
      s >>= 1;
      pos++;
    }
    ans += p[i] - pos;
  }
  while (s != 0) {
    if ((s & 1) == 0) ans++;
    s >>= 1;
  }
  printf("%d\n", ans);
}

### 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>
int N, p[200000];
int s, ans;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", p + i);
  s = 0;
  ans = p[1];
  for (int i = 1, pos = 0; i <= N;) {
    pos = p[i];
    while (i < N && p[i + 1] == p[i]) i++, s++;
    i++;
    s++;
    if (i > N) break;
    while (s && pos < p[i]) {
      if ((s & 1) == 0) ans++;
      s >>= 1;
      pos++;
    }
    ans += p[i] - pos;
  }
  while (s != 0) {
    if ((s & 1) == 0) ans++;
    s >>= 1;
  }
  printf("%d\n", ans);
}

```