#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ull unsigned long long
#define re register
#define il inline
#define inf 0x3f3f3f

int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  std::cin >> n;
  int ans = 0, now = 0;
  while (n--) {
    int tmp;
    std::cin >> tmp;

    int a = log2(tmp);

    if ((1 << a) < tmp)
      if (now > a)
        now = a;
      else
        a++;

    if (a > now)
      ans += a - now;
    now = a;
  }
  std::cout << ans << '\n';
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

There are A slimes lining up in a row. Initially, the sizes of the slimes are all 1.

Snuke can repeatedly perform the following operation.

* Choose a positive even number M. Then, select M consecutive slimes and form M / 2 pairs from those slimes as follows: pair the 1-st and 2-nd of them from the left, the 3-rd and 4-th of them, ..., the (M-1)-th and M-th of them. Combine each pair of slimes into one larger slime. Here, the size of a combined slime is the sum of the individual slimes before combination. The order of the M / 2 combined slimes remain the same as the M / 2 pairs of slimes before combination.



Snuke wants to get to the situation where there are exactly N slimes, and the size of the i-th (1 ≤ i ≤ N) slime from the left is a_i. Find the minimum number of operations required to achieve his goal.

Note that A is not directly given as input. Assume A = a_1 + a_2 + ... + a_N.

Constraints

* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the minimum number of operations required to achieve Snuke's goal.

Examples

Input

2
3 3


Output

2


Input

4
2 1 2 2


Output

2


Input

1
1


Output

0


Input

10
3 1 4 1 5 9 2 6 5 3


Output

10

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ull unsigned long long
#define re register
#define il inline
#define inf 0x3f3f3f

int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  std::cin >> n;
  int ans = 0, now = 0;
  while (n--) {
    int tmp;
    std::cin >> tmp;

    int a = log2(tmp);

    if ((1 << a) < tmp)
      if (now > a)
        now = a;
      else
        a++;

    if (a > now)
      ans += a - now;
    now = a;
  }
  std::cout << ans << '\n';
  return 0;
}

```