#include <bits/stdc++.h>
using namespace std;
const long long int M = 1e9 + 7;
long long int inf = 1e18;
long long int n, m;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int i, j, t, k, x, y, z, N;
  cin >> n;
  long long int a[n];
  map<long long int, long long int> cnt;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  for (auto u : cnt) {
    if (u.second / 2 > 0) {
      cnt[u.first + 1] += u.second / 2;
      if (u.second % 2)
        cnt[u.first] = 1;
      else
        cnt[u.first] = 0;
    }
  }
  long long int mx = -1;
  long long int count = 0;
  for (auto u : cnt) {
    if (u.second) {
      count++;
      mx = max(mx, u.first);
    }
  }
  cout << mx + 1 - count;
  return 0;
}

### Prompt

Your task is to create 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;
const long long int M = 1e9 + 7;
long long int inf = 1e18;
long long int n, m;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int i, j, t, k, x, y, z, N;
  cin >> n;
  long long int a[n];
  map<long long int, long long int> cnt;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  for (auto u : cnt) {
    if (u.second / 2 > 0) {
      cnt[u.first + 1] += u.second / 2;
      if (u.second % 2)
        cnt[u.first] = 1;
      else
        cnt[u.first] = 0;
    }
  }
  long long int mx = -1;
  long long int count = 0;
  for (auto u : cnt) {
    if (u.second) {
      count++;
      mx = max(mx, u.first);
    }
  }
  cout << mx + 1 - count;
  return 0;
}

```