#include <bits/stdc++.h>
using namespace std;
long long A[111111];
int main(void) {
  int n;
  long long res = 0;
  memset(A, 255, sizeof(A));
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  long long cur = 0;
  long long len = 0;
  for (int i = 0; i < n; i++) {
    while (A[i] > cur) {
      if (len == 0) {
        res += (A[i] - cur);
        cur = A[i];
      } else {
        long long c = len - 1;
        if (c % 2) c++;
        res += (c - len + 1);
        len = c / 2;
        cur++;
      }
    }
    len++;
    if (A[i + 1] != A[i]) {
      long long c = len - 1;
      if (c % 2) c++;
      res += (c - len + 1);
      len = c / 2;
      cur++;
    }
  }
  while (len) {
    long long c = len - 1;
    if (c % 2) c++;
    res += (c - len + 1);
    len = c / 2;
    cur++;
  }
  cout << res << endl;
  return 0;
}

### Prompt

Generate 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;
long long A[111111];
int main(void) {
  int n;
  long long res = 0;
  memset(A, 255, sizeof(A));
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  long long cur = 0;
  long long len = 0;
  for (int i = 0; i < n; i++) {
    while (A[i] > cur) {
      if (len == 0) {
        res += (A[i] - cur);
        cur = A[i];
      } else {
        long long c = len - 1;
        if (c % 2) c++;
        res += (c - len + 1);
        len = c / 2;
        cur++;
      }
    }
    len++;
    if (A[i + 1] != A[i]) {
      long long c = len - 1;
      if (c % 2) c++;
      res += (c - len + 1);
      len = c / 2;
      cur++;
    }
  }
  while (len) {
    long long c = len - 1;
    if (c % 2) c++;
    res += (c - len + 1);
    len = c / 2;
    cur++;
  }
  cout << res << endl;
  return 0;
}

```