#include <vector> 
#include <list> 
#include <map> 
#include <set> 
#include <deque> 
#include <queue> 
#include <stack> 
#include <bitset> 
#include <algorithm> 
#include <functional> 
#include <numeric> 
#include <utility> 
#include <sstream> 
#include <iostream> 
#include <iomanip> 
#include <cstdio> 
#include <cmath> 
#include <cstdlib> 
#include <cctype> 
#include <string> 
#include <cstring> 
#include <ctime> 

using namespace std;

#define _int64 long long

int a[110000];
_int64 d[110000][32];

int main()
{
  _int64 ans;
  int i,j,n,pre,b1,tmp1,tmp2,tmp3,tmp4,mi,c0,c1,tmp;
  scanf("%d",&n);
  for (i=0;i<n;i++)
    scanf("%d",&a[i]);
  memset(d,-1,sizeof(d));
  d[0][0]=0;
  for (i=0;i<n;i++)
  {
    tmp=a[i];
    c0=0;c1=0;
    while (tmp!=1)
    {
      c0++;
      if (tmp%2==1) c1=1;
      tmp=(tmp+1)/2;
    }
    if ((c0>0)&&(c1==1))
    {
      c0--;
    }
    //cout<<"c:"<<c0<<" "<<c1<<endl;
    for (j=0;j<32;j++)
      if (d[i][j]!=-1)
      {
        tmp=d[i][j];
        tmp1=j;
        tmp2=c0;
        tmp3=c1;
        tmp4=0;
        if (tmp2>=tmp1)
        {
          tmp4=c0+c1;
        }
        else if (tmp2+tmp3>=tmp1)
        {
          tmp+=(tmp1-tmp2);
          tmp4=tmp2+tmp3-(tmp1-tmp2);
        }
        else
        {
          tmp4=tmp2;
          tmp+=tmp1-tmp2;
        }
        if ((d[i+1][tmp4]==-1)||(tmp<d[i+1][tmp4]))
          d[i+1][tmp4]=tmp;
        //cout<<"4: "<<tmp4<<" "<<tmp<<endl;
      }
  }
  ans=2000000000;
  for (i=0;i<32;i++)
    if (d[n][i]!=-1)
    {
      if (d[n][i]+i<ans) ans=d[n][i]+i;
    }
  printf("%lld\n",ans);
  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 <vector> 
#include <list> 
#include <map> 
#include <set> 
#include <deque> 
#include <queue> 
#include <stack> 
#include <bitset> 
#include <algorithm> 
#include <functional> 
#include <numeric> 
#include <utility> 
#include <sstream> 
#include <iostream> 
#include <iomanip> 
#include <cstdio> 
#include <cmath> 
#include <cstdlib> 
#include <cctype> 
#include <string> 
#include <cstring> 
#include <ctime> 

using namespace std;

#define _int64 long long

int a[110000];
_int64 d[110000][32];

int main()
{
  _int64 ans;
  int i,j,n,pre,b1,tmp1,tmp2,tmp3,tmp4,mi,c0,c1,tmp;
  scanf("%d",&n);
  for (i=0;i<n;i++)
    scanf("%d",&a[i]);
  memset(d,-1,sizeof(d));
  d[0][0]=0;
  for (i=0;i<n;i++)
  {
    tmp=a[i];
    c0=0;c1=0;
    while (tmp!=1)
    {
      c0++;
      if (tmp%2==1) c1=1;
      tmp=(tmp+1)/2;
    }
    if ((c0>0)&&(c1==1))
    {
      c0--;
    }
    //cout<<"c:"<<c0<<" "<<c1<<endl;
    for (j=0;j<32;j++)
      if (d[i][j]!=-1)
      {
        tmp=d[i][j];
        tmp1=j;
        tmp2=c0;
        tmp3=c1;
        tmp4=0;
        if (tmp2>=tmp1)
        {
          tmp4=c0+c1;
        }
        else if (tmp2+tmp3>=tmp1)
        {
          tmp+=(tmp1-tmp2);
          tmp4=tmp2+tmp3-(tmp1-tmp2);
        }
        else
        {
          tmp4=tmp2;
          tmp+=tmp1-tmp2;
        }
        if ((d[i+1][tmp4]==-1)||(tmp<d[i+1][tmp4]))
          d[i+1][tmp4]=tmp;
        //cout<<"4: "<<tmp4<<" "<<tmp<<endl;
      }
  }
  ans=2000000000;
  for (i=0;i<32;i++)
    if (d[n][i]!=-1)
    {
      if (d[n][i]+i<ans) ans=d[n][i]+i;
    }
  printf("%lld\n",ans);
  return 0;
}
```