#include <bits/stdc++.h>
using namespace std;
int main() {
  long long ans = 0, lx, cnt, tp;
  int n;
  scanf("%d", &n);
  lx = -1, cnt = 1;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &tp);
    while (lx < tp && cnt > 1) {
      lx++;
      cnt--;
      if (cnt % 2 == 0) {
        cnt /= 2;
      } else {
        ans++;
        cnt = cnt / 2 + 1;
      }
    }
    if (lx == tp) {
      cnt++;
    } else if (lx < tp) {
      ans += tp - lx - cnt;
      lx = tp;
      cnt = 1;
    }
  }
  while (cnt > 1) {
    cnt--;
    if (cnt % 2 == 0) {
      cnt /= 2;
    } else {
      ans++;
      cnt = cnt / 2 + 1;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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() {
  long long ans = 0, lx, cnt, tp;
  int n;
  scanf("%d", &n);
  lx = -1, cnt = 1;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &tp);
    while (lx < tp && cnt > 1) {
      lx++;
      cnt--;
      if (cnt % 2 == 0) {
        cnt /= 2;
      } else {
        ans++;
        cnt = cnt / 2 + 1;
      }
    }
    if (lx == tp) {
      cnt++;
    } else if (lx < tp) {
      ans += tp - lx - cnt;
      lx = tp;
      cnt = 1;
    }
  }
  while (cnt > 1) {
    cnt--;
    if (cnt % 2 == 0) {
      cnt /= 2;
    } else {
      ans++;
      cnt = cnt / 2 + 1;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```