#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
map<int, int> cnt;
vector<int> vctr;
int main() {
  int n, a1;
  cin >> n;
  for (int i = 0; i < n; ++i, ++cnt[a1], vctr.push_back(a1)) cin >> a1;
  sort(vctr.begin(), vctr.end());
  vctr.resize(unique(vctr.begin(), vctr.end()) - vctr.begin());
  reverse(vctr.begin(), vctr.end());
  int sum = 0, mx = -1;
  while (!vctr.empty()) {
    int now = vctr.back();
    while (vctr.back() == now) vctr.pop_back();
    cnt[now + 1] += cnt[now] / 2;
    cnt[now] %= 2;
    sum += cnt[now];
    if (cnt[now + 1]) vctr.push_back(now + 1);
    mx = max(mx, now);
  }
  cout << mx + 1 - sum;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 int MAXN = 1e5 + 10;
map<int, int> cnt;
vector<int> vctr;
int main() {
  int n, a1;
  cin >> n;
  for (int i = 0; i < n; ++i, ++cnt[a1], vctr.push_back(a1)) cin >> a1;
  sort(vctr.begin(), vctr.end());
  vctr.resize(unique(vctr.begin(), vctr.end()) - vctr.begin());
  reverse(vctr.begin(), vctr.end());
  int sum = 0, mx = -1;
  while (!vctr.empty()) {
    int now = vctr.back();
    while (vctr.back() == now) vctr.pop_back();
    cnt[now + 1] += cnt[now] / 2;
    cnt[now] %= 2;
    sum += cnt[now];
    if (cnt[now + 1]) vctr.push_back(now + 1);
    mx = max(mx, now);
  }
  cout << mx + 1 - sum;
  return 0;
}

```