#include <bits/stdc++.h>
using namespace std;
map<int, int> mp;
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    mp[x]++;
  }
  int last = -1;
  int ans = 0;
  map<int, int>::iterator it = mp.begin(), itr = mp.end();
  while (it != itr) {
    int cur = (*it).first;
    if (((*it).second > 1)) {
      mp[cur + 1] += (*it).second / 2;
      mp[cur] = (*it).second % 2;
    }
    if (mp[cur]) {
      ans += cur - last - 1;
      last = cur;
    }
    it++;
  }
  printf("%d\n", ans);
}

### Prompt

Generate 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;
map<int, int> mp;
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    mp[x]++;
  }
  int last = -1;
  int ans = 0;
  map<int, int>::iterator it = mp.begin(), itr = mp.end();
  while (it != itr) {
    int cur = (*it).first;
    if (((*it).second > 1)) {
      mp[cur + 1] += (*it).second / 2;
      mp[cur] = (*it).second % 2;
    }
    if (mp[cur]) {
      ans += cur - last - 1;
      last = cur;
    }
    it++;
  }
  printf("%d\n", ans);
}

```