#include <bits/stdc++.h>
using namespace std;
map<int, int> mp;
vector<int> v;
int i, j, k, l, m, n, mx, ans;
void solve(int x) {
  if (!mp[x]) return;
  mx = max(mx, x);
  if (mp[x] % 2) ans++;
  if (mp[x] > 1) {
    mp[x + 1] += mp[x] / 2;
    mp[x] %= 2;
    solve(x + 1);
  }
}
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> l;
    mp[l]++;
    if (i == 0) {
      v.push_back(l);
      continue;
    }
    if (l != v.back()) v.push_back(l);
  }
  for (i = 0; i < v.size(); i++) solve(v[i]);
  cout << mx - ans + 1 << endl;
}

### 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;
map<int, int> mp;
vector<int> v;
int i, j, k, l, m, n, mx, ans;
void solve(int x) {
  if (!mp[x]) return;
  mx = max(mx, x);
  if (mp[x] % 2) ans++;
  if (mp[x] > 1) {
    mp[x + 1] += mp[x] / 2;
    mp[x] %= 2;
    solve(x + 1);
  }
}
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> l;
    mp[l]++;
    if (i == 0) {
      v.push_back(l);
      continue;
    }
    if (l != v.back()) v.push_back(l);
  }
  for (i = 0; i < v.size(); i++) solve(v[i]);
  cout << mx - ans + 1 << endl;
}

```