#include <bits/stdc++.h>
using namespace std;
priority_queue<int> Q;
int main() {
  int n, val;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &val);
    Q.push(-val);
  }
  int pos = -1, c = 0, ans = 0;
  while (!Q.empty()) {
    if (c == 0) {
      ans += -Q.top() - pos - 1;
      pos = -Q.top();
    }
    while (!Q.empty() && -Q.top() == pos) {
      c++;
      Q.pop();
    }
    if (c % 2 == 0) {
      ans++;
    }
    c /= 2;
    if (c) pos++;
  }
  while (c) {
    if (c % 2 == 0) ans++;
    c /= 2;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
priority_queue<int> Q;
int main() {
  int n, val;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &val);
    Q.push(-val);
  }
  int pos = -1, c = 0, ans = 0;
  while (!Q.empty()) {
    if (c == 0) {
      ans += -Q.top() - pos - 1;
      pos = -Q.top();
    }
    while (!Q.empty() && -Q.top() == pos) {
      c++;
      Q.pop();
    }
    if (c % 2 == 0) {
      ans++;
    }
    c /= 2;
    if (c) pos++;
  }
  while (c) {
    if (c % 2 == 0) ans++;
    c /= 2;
  }
  cout << ans << endl;
  return 0;
}

```