#include<bits/stdc++.h>

using namespace std;

const int INF = 1 << 30;

int main()
{
  int N, A[100000];

  cin >> N;
  for(int i = 0; i < N; i++) {
    cin >> A[i];
  }

  int ret = 0;
  vector< int > dp(31, INF);
  dp[0] = 0;
  for(int i = 0; i < N; i++) {
    int z = A[i], x = 0;
    while(z > 1) z = (z + 1) >> 1, ++x;
    vector< int > nt(31, INF);
    if(__builtin_popcount(A[i]) == 1) {
      for(int j = 0; j < 31; j++) nt[x] = min(nt[x], dp[j] + max(0, x - j));
    } else {
      for(int j = 0; j < 31; j++) nt[x] = min(nt[x], dp[j] + max(0, x - j - 1) + 1);
      for(int j = 1; j < 31; j++) nt[x - 1] = min(nt[x - 1], dp[j] + max(0, x - j));
    }
    nt.swap(dp);
  }
  cout << *min_element(begin(dp), end(dp)) << endl;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;

const int INF = 1 << 30;

int main()
{
  int N, A[100000];

  cin >> N;
  for(int i = 0; i < N; i++) {
    cin >> A[i];
  }

  int ret = 0;
  vector< int > dp(31, INF);
  dp[0] = 0;
  for(int i = 0; i < N; i++) {
    int z = A[i], x = 0;
    while(z > 1) z = (z + 1) >> 1, ++x;
    vector< int > nt(31, INF);
    if(__builtin_popcount(A[i]) == 1) {
      for(int j = 0; j < 31; j++) nt[x] = min(nt[x], dp[j] + max(0, x - j));
    } else {
      for(int j = 0; j < 31; j++) nt[x] = min(nt[x], dp[j] + max(0, x - j - 1) + 1);
      for(int j = 1; j < 31; j++) nt[x - 1] = min(nt[x - 1], dp[j] + max(0, x - j));
    }
    nt.swap(dp);
  }
  cout << *min_element(begin(dp), end(dp)) << endl;
}
```