#include <bits/stdc++.h>
using namespace std;
int main() {
  priority_queue<long long, vector<long long>, greater<long long> > q;
  map<long long, long long> mp;
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) {
      long long x;
      scanf("%I64d", &x);
      mp[x]++;
    }
  }
  map<long long, long long>::iterator it;
  long long cnt = 0;
  for (it = mp.begin(); it != mp.end(); it++) {
    long long a = it->first, b = it->second;
    int x = b - 1;
    if (x == 0) continue;
    if (x % 2 != 0) {
      cnt++;
      x++;
    }
    mp[a + 1] += x / 2;
  }
  printf("%I64d\n", mp.rbegin()->first + 1 - mp.size() + cnt);
  return 0;
}

### Prompt

Your challenge is to write 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;
int main() {
  priority_queue<long long, vector<long long>, greater<long long> > q;
  map<long long, long long> mp;
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) {
      long long x;
      scanf("%I64d", &x);
      mp[x]++;
    }
  }
  map<long long, long long>::iterator it;
  long long cnt = 0;
  for (it = mp.begin(); it != mp.end(); it++) {
    long long a = it->first, b = it->second;
    int x = b - 1;
    if (x == 0) continue;
    if (x % 2 != 0) {
      cnt++;
      x++;
    }
    mp[a + 1] += x / 2;
  }
  printf("%I64d\n", mp.rbegin()->first + 1 - mp.size() + cnt);
  return 0;
}

```