#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
long power(long x, long m){
    if(m == 0) return 1;
    long y = power(x, m/2);
    y = y * y;
    if(m&1) y = y * x;
    return y;
}
long l2(long z){
    long a=1,e=0;
    while(a<z){
        a*=2;
        e++;
    }
    return e;
}
long long n,i,c,p,s,lg;
int main(){
    cin>>n;
    cin>>c;
    p=l2(c);
    s=s+p;
    for(i=1;i<n;i++){
        cin>>c;
        lg=l2(c);
        if(c==1){
            p=0;
        }
        else if(p>=lg){
            if(c-power(2,lg)==0){
                p=lg;
            }
            else{
                p=lg-1;
            }
        }
        else{
            s=s+lg-p;
            p=lg;
        }
    }
    cout<<s;
}
### 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 <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
long power(long x, long m){
    if(m == 0) return 1;
    long y = power(x, m/2);
    y = y * y;
    if(m&1) y = y * x;
    return y;
}
long l2(long z){
    long a=1,e=0;
    while(a<z){
        a*=2;
        e++;
    }
    return e;
}
long long n,i,c,p,s,lg;
int main(){
    cin>>n;
    cin>>c;
    p=l2(c);
    s=s+p;
    for(i=1;i<n;i++){
        cin>>c;
        lg=l2(c);
        if(c==1){
            p=0;
        }
        else if(p>=lg){
            if(c-power(2,lg)==0){
                p=lg;
            }
            else{
                p=lg-1;
            }
        }
        else{
            s=s+lg-p;
            p=lg;
        }
    }
    cout<<s;
}
```