#include <bits/stdc++.h>

using namespace std;

bool power2(int n) {
  if (n == 1) return true;
  if (n & 1) return false;
  return power2(n / 2);
}

int main() {
  int n;
  scanf("%d", &n);
  vector<int> x(n);
  vector<bool> isPower2(n);
  int result = 0;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    int k = ceil(log2(a));
    x[i] = k;
    result += k;
    isPower2[i] = power2(a);
  }
  vector<int> savedLeft(n), savedRight(n);
  for (int i = 1; i < n; i++) {
    int beforeBig = x[i - 1];
    int beforeSmall = x[i - 1] - !isPower2[i - 1];

    int curBig = x[i];
    int curSmall = x[i] - !isPower2[i];

    savedLeft[i] = max(
        savedLeft[i - 1] + min(beforeBig, curSmall),
        savedRight[i - 1] + min(beforeSmall, curSmall));
    savedRight[i] = max(
        savedLeft[i - 1] + min(beforeBig, curBig),
        savedRight[i - 1] + min(beforeSmall, curBig));
  }
  printf("%d\n", result - max(savedLeft[n-1], savedRight[n-1]));
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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;

bool power2(int n) {
  if (n == 1) return true;
  if (n & 1) return false;
  return power2(n / 2);
}

int main() {
  int n;
  scanf("%d", &n);
  vector<int> x(n);
  vector<bool> isPower2(n);
  int result = 0;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    int k = ceil(log2(a));
    x[i] = k;
    result += k;
    isPower2[i] = power2(a);
  }
  vector<int> savedLeft(n), savedRight(n);
  for (int i = 1; i < n; i++) {
    int beforeBig = x[i - 1];
    int beforeSmall = x[i - 1] - !isPower2[i - 1];

    int curBig = x[i];
    int curSmall = x[i] - !isPower2[i];

    savedLeft[i] = max(
        savedLeft[i - 1] + min(beforeBig, curSmall),
        savedRight[i - 1] + min(beforeSmall, curSmall));
    savedRight[i] = max(
        savedLeft[i - 1] + min(beforeBig, curBig),
        savedRight[i - 1] + min(beforeSmall, curBig));
  }
  printf("%d\n", result - max(savedLeft[n-1], savedRight[n-1]));
  return 0;
}

```