#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  map<long long, long long> m;
  int v;
  for (int i = 0; i < n; i++) {
    scanf("%d", &v);
    m[v]++;
  }
  for (auto it = m.begin(); it != m.end(); it++) {
    if (it->second > 1) m[it->first + 1] += it->second / 2;
    it->second %= 2;
  }
  long long ans = 0;
  vector<long long> w;
  for (auto it = m.begin(); it != m.end(); it++)
    if (it->second) w.push_back(it->first);
  long long pre = -1;
  for (int i = 0; i < w.size(); i++) {
    ans += w[i] - pre - 1;
    pre = w[i];
  }
  cout << ans << endl;
}

### Prompt

In Cpp, your task is to solve 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() {
  int n;
  scanf("%d", &n);
  map<long long, long long> m;
  int v;
  for (int i = 0; i < n; i++) {
    scanf("%d", &v);
    m[v]++;
  }
  for (auto it = m.begin(); it != m.end(); it++) {
    if (it->second > 1) m[it->first + 1] += it->second / 2;
    it->second %= 2;
  }
  long long ans = 0;
  vector<long long> w;
  for (auto it = m.begin(); it != m.end(); it++)
    if (it->second) w.push_back(it->first);
  long long pre = -1;
  for (int i = 0; i < w.size(); i++) {
    ans += w[i] - pre - 1;
    pre = w[i];
  }
  cout << ans << endl;
}

```