#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
int main() {
  long long int n;
  cin >> n;
  vector<long long int> op(n);
  map<long long int, long long int> count;
  for (long long int i = 0; i < n; i++) {
    scanf("%I64d", &op[i]);
    count[op[i]]++;
  }
  long long int ans = 0, last = -1;
  for (map<long long int, long long int>::iterator it = count.begin();
       it != count.end(); it++) {
    long long int to = 0, two = 1;
    while (it->second > two) {
      to++;
      two *= 2;
    }
    while (to > 0) {
      if (it->second >= two) {
        count[it->first + to]++;
        it->second -= two;
      }
      two /= 2;
      to--;
    }
    ans += it->first - last - 1;
    last = it->first;
    ans += (1 - it->second);
  }
  cout << ans;
}

### 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>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
int main() {
  long long int n;
  cin >> n;
  vector<long long int> op(n);
  map<long long int, long long int> count;
  for (long long int i = 0; i < n; i++) {
    scanf("%I64d", &op[i]);
    count[op[i]]++;
  }
  long long int ans = 0, last = -1;
  for (map<long long int, long long int>::iterator it = count.begin();
       it != count.end(); it++) {
    long long int to = 0, two = 1;
    while (it->second > two) {
      to++;
      two *= 2;
    }
    while (to > 0) {
      if (it->second >= two) {
        count[it->first + to]++;
        it->second -= two;
      }
      two /= 2;
      to--;
    }
    ans += it->first - last - 1;
    last = it->first;
    ans += (1 - it->second);
  }
  cout << ans;
}

```