#include <bits/stdc++.h>
using namespace std;
const int Inf = 1000000000;
const int N = 100010;
long long a[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  long long now = 0, ans = 0, k = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == k) {
      now++;
      continue;
    }
    while (k < a[i] && now != 0) {
      if (now % 2 == 0) ans++;
      now /= 2;
      k++;
    }
    ans += a[i] - k;
    k = a[i];
    now++;
  }
  while (now) {
    if (now % 2 == 0) ans++;
    now /= 2;
    k++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int Inf = 1000000000;
const int N = 100010;
long long a[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  long long now = 0, ans = 0, k = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == k) {
      now++;
      continue;
    }
    while (k < a[i] && now != 0) {
      if (now % 2 == 0) ans++;
      now /= 2;
      k++;
    }
    ans += a[i] - k;
    k = a[i];
    now++;
  }
  while (now) {
    if (now % 2 == 0) ans++;
    now /= 2;
    k++;
  }
  cout << ans << endl;
  return 0;
}

```