#include <cstdio>
#include <algorithm>

int main() {
  int n;
  scanf("%d", &n);
  int dp0 = 0;
  int dp1 = 0;
  int prevLow = 0;
  int prevHigh = 0;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    int twos = 0;
    while ((1LL << (twos + 1)) <= x) {
      ++twos;
    }
    int low = twos;
    int high = twos;
    if ((1 << twos) != x) {
      high++;
    }
    int next0 = std::min(dp0 + std::max(0, high - prevLow), dp1 + std::max(0, high - prevHigh));
    int next1 = std::min(dp0 + std::max(0, low - prevLow) + high - low, dp1 + std::max(0, low - prevHigh) + high - low);
    dp0 = next0;
    dp1 = next1;
    prevLow = low;
    prevHigh = high; 
  }
  printf("%d\n", std::min(dp0, dp1));
}

### Prompt

Generate a cpp solution 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 <cstdio>
#include <algorithm>

int main() {
  int n;
  scanf("%d", &n);
  int dp0 = 0;
  int dp1 = 0;
  int prevLow = 0;
  int prevHigh = 0;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    int twos = 0;
    while ((1LL << (twos + 1)) <= x) {
      ++twos;
    }
    int low = twos;
    int high = twos;
    if ((1 << twos) != x) {
      high++;
    }
    int next0 = std::min(dp0 + std::max(0, high - prevLow), dp1 + std::max(0, high - prevHigh));
    int next1 = std::min(dp0 + std::max(0, low - prevLow) + high - low, dp1 + std::max(0, low - prevHigh) + high - low);
    dp0 = next0;
    dp1 = next1;
    prevLow = low;
    prevHigh = high; 
  }
  printf("%d\n", std::min(dp0, dp1));
}

```