#include<bits/stdc++.h>
#define N 100005
#define ll long long
using namespace std;
int n;
int a[N],b[N],c[N];
int main()
{
    cin>>n;
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        int tmp=a[i];
        while(tmp)
        {
            b[i]++;
            tmp>>=1;
        }
        b[i]--;
        if((a[i]&-a[i])!=a[i])c[i]=1;
        ans+=b[i]+c[i];
    }
    int lst=0;
    for(int i=1;i<=n;i++)
    {
        int tmp=min(b[i],lst);
        ans-=tmp;lst-=tmp;
        if(lst>0&&c[i]>0)
        {
            ans--;c[i]--;
        }
        lst=b[i]+c[i];
    }
    cout<<ans<<endl;
    return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 N 100005
#define ll long long
using namespace std;
int n;
int a[N],b[N],c[N];
int main()
{
    cin>>n;
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        int tmp=a[i];
        while(tmp)
        {
            b[i]++;
            tmp>>=1;
        }
        b[i]--;
        if((a[i]&-a[i])!=a[i])c[i]=1;
        ans+=b[i]+c[i];
    }
    int lst=0;
    for(int i=1;i<=n;i++)
    {
        int tmp=min(b[i],lst);
        ans-=tmp;lst-=tmp;
        if(lst>0&&c[i]>0)
        {
            ans--;c[i]--;
        }
        lst=b[i]+c[i];
    }
    cout<<ans<<endl;
    return 0;
}

```