#include<iostream>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
using namespace std;

int n;
int a[100011];
int r[100011][2];
int p[100011];
int ans;
int dp[100011][2];

int main()
{
    int i,j,k;
    
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        int xx=0,yy=a[i];
        while(yy>1)
        {
            yy>>=1;
            xx++;
        }
        if((a[i]^(1<<xx))!=0)
        {
            r[i][0]=xx+1;
            r[i][1]=xx;
            p[i]=xx+1;
        }
        else
        {
            r[i][0]=r[i][1]=p[i]=xx;
        }
    }
    
    memset(dp,-1,sizeof(dp));
    dp[0][0]=dp[0][1]=0;
    
    for(i=0;i<n;i++)
    {
        for(j=0;j<=1;j++)
        {
            if(dp[i][j]!=-1)
            {
                for(k=0;k<=1;k++)
                {
                    dp[i+1][k]=max(dp[i+1][k],dp[i][j]+min(r[i][1-j],r[i+1][k]));
                }
            }
        }
    }
    ans=0;
    for(i=1;i<=n;i++)
    {
        ans+=p[i];
    }
    
    ans-=max(dp[n][0],dp[n][1]);
    
    cout<<ans<<endl;
    
    return 0;
}

### Prompt

Create a solution in CPP for 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<iostream>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
using namespace std;

int n;
int a[100011];
int r[100011][2];
int p[100011];
int ans;
int dp[100011][2];

int main()
{
    int i,j,k;
    
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        int xx=0,yy=a[i];
        while(yy>1)
        {
            yy>>=1;
            xx++;
        }
        if((a[i]^(1<<xx))!=0)
        {
            r[i][0]=xx+1;
            r[i][1]=xx;
            p[i]=xx+1;
        }
        else
        {
            r[i][0]=r[i][1]=p[i]=xx;
        }
    }
    
    memset(dp,-1,sizeof(dp));
    dp[0][0]=dp[0][1]=0;
    
    for(i=0;i<n;i++)
    {
        for(j=0;j<=1;j++)
        {
            if(dp[i][j]!=-1)
            {
                for(k=0;k<=1;k++)
                {
                    dp[i+1][k]=max(dp[i+1][k],dp[i][j]+min(r[i][1-j],r[i+1][k]));
                }
            }
        }
    }
    ans=0;
    for(i=1;i<=n;i++)
    {
        ans+=p[i];
    }
    
    ans-=max(dp[n][0],dp[n][1]);
    
    cout<<ans<<endl;
    
    return 0;
}

```