#include<iostream>
#include<cstdio>
#include<ctype.h>
#include<cmath>
using namespace std;
inline int read(){
    int x=0,f=0;char ch=getchar();
    while(!isdigit(ch))f|=ch=='-',ch=getchar();
    while(isdigit(ch))x=x*10+(ch^48),ch=getchar();
    return f?-x:x;
}
int main(){
    int n=read(),ans=0,last=0;
    for(int i=1;i<=n;++i){
    	int x=read(),a=log2(x);
        if((1<<a)<x && last<=a)a++;
        if(a>last)ans+=a-last;
        last=a;
    }
    printf("%d", ans);
    return 0;
}
 
### Prompt

Generate a cpp solution 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<iostream>
#include<cstdio>
#include<ctype.h>
#include<cmath>
using namespace std;
inline int read(){
    int x=0,f=0;char ch=getchar();
    while(!isdigit(ch))f|=ch=='-',ch=getchar();
    while(isdigit(ch))x=x*10+(ch^48),ch=getchar();
    return f?-x:x;
}
int main(){
    int n=read(),ans=0,last=0;
    for(int i=1;i<=n;++i){
    	int x=read(),a=log2(x);
        if((1<<a)<x && last<=a)a++;
        if(a>last)ans+=a-last;
        last=a;
    }
    printf("%d", ans);
    return 0;
}
 
```