#include <bits/stdc++.h>
using namespace std;
int main() {
  int n_case;
  cin >> n_case;
  vector<long long> vec;
  long long ans = 0;
  for (int i = 0; i < n_case; i++) {
    long long tmp;
    cin >> tmp;
    vec.push_back(tmp);
  }
  long long curbit = 0;
  long long carry = 0;
  for (int i = 0; i < n_case; i++) {
    long long cur = vec[i];
    if (curbit < cur) {
      while (carry != 0 && curbit != cur) {
        curbit++;
        if (carry % 2 == 0) ans++;
        carry /= 2;
      }
      if (curbit < cur) {
        ans += cur - curbit;
      }
    }
    curbit = cur;
    while (i < n_case && vec[i] == cur) {
      i++;
      carry++;
    }
    if (carry % 2 == 0) ans++;
    carry /= 2;
    curbit++;
    i--;
  }
  while (carry != 0) {
    if (carry % 2 == 0) ans++;
    carry /= 2;
  }
  cout << ans << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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() {
  int n_case;
  cin >> n_case;
  vector<long long> vec;
  long long ans = 0;
  for (int i = 0; i < n_case; i++) {
    long long tmp;
    cin >> tmp;
    vec.push_back(tmp);
  }
  long long curbit = 0;
  long long carry = 0;
  for (int i = 0; i < n_case; i++) {
    long long cur = vec[i];
    if (curbit < cur) {
      while (carry != 0 && curbit != cur) {
        curbit++;
        if (carry % 2 == 0) ans++;
        carry /= 2;
      }
      if (curbit < cur) {
        ans += cur - curbit;
      }
    }
    curbit = cur;
    while (i < n_case && vec[i] == cur) {
      i++;
      carry++;
    }
    if (carry % 2 == 0) ans++;
    carry /= 2;
    curbit++;
    i--;
  }
  while (carry != 0) {
    if (carry % 2 == 0) ans++;
    carry /= 2;
  }
  cout << ans << endl;
}

```