#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline int popcount(T t) {
  if (std::numeric_limits<T>::digits <=
      std::numeric_limits<unsigned int>::digits) {
    return __builtin_popcount(t);
  } else {
    return __builtin_popcountll(t);
  }
}
const long double EPS = 1e-8;
const long double PI = acosl(0.0) * 2;
void solve() {
  long long ans = 0;
  int n;
  cin >> n;
  long long last = 0, cnt = 0;
  for (int i = 0; i < int(n); ++i) {
    long long x;
    cin >> x;
    while (last < x && cnt > 0) {
      if (cnt % 2 == 0) ++ans;
      cnt /= 2;
      ++last;
    }
    ans += x - last;
    last = x;
    cnt += 1;
  }
  while (cnt > 0) {
    if (cnt % 2 == 0) ++ans;
    cnt /= 2;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}

### Prompt

Develop a solution in cpp 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;
template <typename T>
inline int popcount(T t) {
  if (std::numeric_limits<T>::digits <=
      std::numeric_limits<unsigned int>::digits) {
    return __builtin_popcount(t);
  } else {
    return __builtin_popcountll(t);
  }
}
const long double EPS = 1e-8;
const long double PI = acosl(0.0) * 2;
void solve() {
  long long ans = 0;
  int n;
  cin >> n;
  long long last = 0, cnt = 0;
  for (int i = 0; i < int(n); ++i) {
    long long x;
    cin >> x;
    while (last < x && cnt > 0) {
      if (cnt % 2 == 0) ++ans;
      cnt /= 2;
      ++last;
    }
    ans += x - last;
    last = x;
    cnt += 1;
  }
  while (cnt > 0) {
    if (cnt % 2 == 0) ++ans;
    cnt /= 2;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}

```