#include <bits/stdc++.h>

#define FI(i,a,b) for(int i=(a);i<=(b);i++)
#define FD(i,a,b) for(int i=(a);i>=(b);i--)

#define LL long long
#define Ldouble long double

using namespace std;

int lev[100005], good[100005];
int n, ans;

int main(){
  scanf("%d", &n);
  FI(i, 1, n){
    int x;
    scanf("%d", &x);
    FI(j, 0, 30){
      if(x <= (1 << j)){
        if(x == (1 << j)) good[i] = true;
        lev[i] = j;
        break;
      }
    }
  }
  FD(i, 30, 1){
    //level i
    int cumu = 0;
    FI(j, 1, n){
      if(lev[j] == i){
        if(!good[j]){
          if(cumu == -1){
            ans++;
            cumu = 0;
          }
          else if(cumu == -2) cumu = 0;
          else cumu = -1;
          
          good[j] = 1;
        }
        else{
          if(cumu == -1){}
          else if(cumu == 0){
            ans++;
            cumu = -2;
          }
        }
        lev[j]--;
      }
      else{
        if(cumu == -1) ans++;
        cumu = 0;
      }
    }
    if(cumu == -1) ans++;
    //printf("%d: %d\n", i, ans);
    //if(i <= 10){
    //  FI(j, 1, n) printf("%d[%d], ", lev[j], good[j]);
    //  printf("\n");
    //}
  }
  printf("%d\n", ans);
  return 0;
}
### 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>

#define FI(i,a,b) for(int i=(a);i<=(b);i++)
#define FD(i,a,b) for(int i=(a);i>=(b);i--)

#define LL long long
#define Ldouble long double

using namespace std;

int lev[100005], good[100005];
int n, ans;

int main(){
  scanf("%d", &n);
  FI(i, 1, n){
    int x;
    scanf("%d", &x);
    FI(j, 0, 30){
      if(x <= (1 << j)){
        if(x == (1 << j)) good[i] = true;
        lev[i] = j;
        break;
      }
    }
  }
  FD(i, 30, 1){
    //level i
    int cumu = 0;
    FI(j, 1, n){
      if(lev[j] == i){
        if(!good[j]){
          if(cumu == -1){
            ans++;
            cumu = 0;
          }
          else if(cumu == -2) cumu = 0;
          else cumu = -1;
          
          good[j] = 1;
        }
        else{
          if(cumu == -1){}
          else if(cumu == 0){
            ans++;
            cumu = -2;
          }
        }
        lev[j]--;
      }
      else{
        if(cumu == -1) ans++;
        cumu = 0;
      }
    }
    if(cumu == -1) ans++;
    //printf("%d: %d\n", i, ans);
    //if(i <= 10){
    //  FI(j, 1, n) printf("%d[%d], ", lev[j], good[j]);
    //  printf("\n");
    //}
  }
  printf("%d\n", ans);
  return 0;
}
```