#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=100010;
int a[N],f[N][31];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline int calc(int x) {
    if(x==1) return 0;
    return 1+calc((x+1)>>1);
}
int main() {
    int n,x,y;
    cin>>n;
    memset(f,127/3,sizeof(f));
    f[0][0]=0;
    for(int i=1;i<=n;i++) {
	x=gi(),y=calc(x);
	if((1<<y)==x) x=y;
	else x=y-1;
	for(int j=0;j<31;j++) {
	    f[i][x]=min(f[i][x],f[i-1][j]+max(0,y-j));
	    f[i][y]=min(f[i][y],f[i-1][j]+max(0,x-j)+y-x);
	}
    }
    cout<<min(f[n][x],f[n][y]);
    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 <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=100010;
int a[N],f[N][31];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline int calc(int x) {
    if(x==1) return 0;
    return 1+calc((x+1)>>1);
}
int main() {
    int n,x,y;
    cin>>n;
    memset(f,127/3,sizeof(f));
    f[0][0]=0;
    for(int i=1;i<=n;i++) {
	x=gi(),y=calc(x);
	if((1<<y)==x) x=y;
	else x=y-1;
	for(int j=0;j<31;j++) {
	    f[i][x]=min(f[i][x],f[i-1][j]+max(0,y-j));
	    f[i][y]=min(f[i][y],f[i-1][j]+max(0,x-j)+y-x);
	}
    }
    cout<<min(f[n][x],f[n][y]);
    return 0;
}
```