#include <bits/stdc++.h>
using namespace std;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ryo;
long long int diff = 0;
int main() {
  long long int n, i, x;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    ryo.push(x);
  }
  long long int sign;
  while (ryo.size() > 0) {
    sign = ryo.top();
    ryo.pop();
    long long int topop = ryo.size(), amount = 1;
    for (i = 0; i < topop; i++) {
      if (ryo.top() == sign) {
        ryo.pop();
        amount++;
      } else
        break;
    }
    if (amount % 2 == 1) diff++;
    for (i = 1; i <= amount / 2; i++) ryo.push(sign + 1);
  }
  cout << sign - diff + 1 << endl;
}

### Prompt

Your task is to create 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;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ryo;
long long int diff = 0;
int main() {
  long long int n, i, x;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    ryo.push(x);
  }
  long long int sign;
  while (ryo.size() > 0) {
    sign = ryo.top();
    ryo.pop();
    long long int topop = ryo.size(), amount = 1;
    for (i = 0; i < topop; i++) {
      if (ryo.top() == sign) {
        ryo.pop();
        amount++;
      } else
        break;
    }
    if (amount % 2 == 1) diff++;
    for (i = 1; i <= amount / 2; i++) ryo.push(sign + 1);
  }
  cout << sign - diff + 1 << endl;
}

```