#include <bits/stdc++.h>
using namespace std;

int main()
{
  int n;
  scanf("%d",&n);
  static long long a[100000];
  for(int i=0;i<n;i++){
    scanf("%lld",a+i);
  }
  int c=0,ans=0;
  for(int i=0;i<n;i++){
    int t=0;
    while(a[i]>(1ll<<t)){
      t++;
    }
    //printf("%d %d\n",c,t);
    if(a[i]%(1ll<<c)==0){
      ans+=t-c;
      c=t;
    }
    else if(a[i]==(1ll<<t)){
      c=t;
    }
    else if(c<=t-1){
      ans+=t-c;
      c=t;
    }
    else{
      c=t-1;
    }
  }
  printf("%d\n",ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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;

int main()
{
  int n;
  scanf("%d",&n);
  static long long a[100000];
  for(int i=0;i<n;i++){
    scanf("%lld",a+i);
  }
  int c=0,ans=0;
  for(int i=0;i<n;i++){
    int t=0;
    while(a[i]>(1ll<<t)){
      t++;
    }
    //printf("%d %d\n",c,t);
    if(a[i]%(1ll<<c)==0){
      ans+=t-c;
      c=t;
    }
    else if(a[i]==(1ll<<t)){
      c=t;
    }
    else if(c<=t-1){
      ans+=t-c;
      c=t;
    }
    else{
      c=t-1;
    }
  }
  printf("%d\n",ans);
  return 0;
}

```