#include <bits/stdc++.h>
using namespace std;
const long long INF = numeric_limits<long long>::max() - 1;
priority_queue<long, vector<long>, greater<long> > q;
int main() {
  ios_base::sync_with_stdio(false);
  long n, a, num = 0, mx = 0;
  cin >> n;
  for (int i = 0; i < (int)n; i++) cin >> a, q.push(a);
  while (!q.empty()) {
    long m = 0, prev = q.top();
    while (!q.empty() && prev == q.top()) {
      m++;
      q.pop();
    }
    for (int i = 0; m; i++, m >>= 1) {
      if (!(m & 1)) continue;
      if (i == 0)
        num++, mx = max(mx, prev);
      else
        q.push(prev + i);
    }
  }
  cout << mx - num + 1 << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
const long long INF = numeric_limits<long long>::max() - 1;
priority_queue<long, vector<long>, greater<long> > q;
int main() {
  ios_base::sync_with_stdio(false);
  long n, a, num = 0, mx = 0;
  cin >> n;
  for (int i = 0; i < (int)n; i++) cin >> a, q.push(a);
  while (!q.empty()) {
    long m = 0, prev = q.top();
    while (!q.empty() && prev == q.top()) {
      m++;
      q.pop();
    }
    for (int i = 0; m; i++, m >>= 1) {
      if (!(m & 1)) continue;
      if (i == 0)
        num++, mx = max(mx, prev);
      else
        q.push(prev + i);
    }
  }
  cout << mx - num + 1 << endl;
  return 0;
}

```