#include<algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include  <stdio.h>
#include   <math.h>
#include   <time.h>
#include   <vector>
#include   <bitset>
#include    <queue>
#include      <set>
#include      <map>
using namespace std;

const int N=100005;

int n,a[N],f[N][32];

int main()
{
	memset(f,1,sizeof f);
	cin>>n;
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	f[0][0]=0;
	for(int i=0;i<n;i++)
	{
		int Num0=0,Num1=0,sz=a[i+1];
		while(sz>1)
		{
			if(sz%2==0)
				Num0++,sz/=2;
			else if(sz%2==1)
				Num0++,Num1++,sz=sz/2+1;
		}
		if(Num1)
			Num1=1,Num0--;
		for(int j=0;j<=30;j++)
		{
			if(Num0>=j)
				f[i+1][Num0+Num1]=min(f[i+1][Num0+Num1],f[i][j]+max(Num0+Num1-j,0));
			else if(Num1+Num0>=j)
				f[i+1][Num0+Num0-j+Num1]=min(f[i+1][Num0+Num0-j+Num1],f[i][j]+max(Num0+Num1-j,0));
			else
				f[i+1][Num0]=min(f[i+1][Num0],f[i][j]+max(Num0+Num1-j,0));
		}
	}
	int Ans=1e9;
	for(int j=0;j<=30;j++)
		Ans=min(Ans,f[n][j]);
	cout<<Ans<<endl;
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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<algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include  <stdio.h>
#include   <math.h>
#include   <time.h>
#include   <vector>
#include   <bitset>
#include    <queue>
#include      <set>
#include      <map>
using namespace std;

const int N=100005;

int n,a[N],f[N][32];

int main()
{
	memset(f,1,sizeof f);
	cin>>n;
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	f[0][0]=0;
	for(int i=0;i<n;i++)
	{
		int Num0=0,Num1=0,sz=a[i+1];
		while(sz>1)
		{
			if(sz%2==0)
				Num0++,sz/=2;
			else if(sz%2==1)
				Num0++,Num1++,sz=sz/2+1;
		}
		if(Num1)
			Num1=1,Num0--;
		for(int j=0;j<=30;j++)
		{
			if(Num0>=j)
				f[i+1][Num0+Num1]=min(f[i+1][Num0+Num1],f[i][j]+max(Num0+Num1-j,0));
			else if(Num1+Num0>=j)
				f[i+1][Num0+Num0-j+Num1]=min(f[i+1][Num0+Num0-j+Num1],f[i][j]+max(Num0+Num1-j,0));
			else
				f[i+1][Num0]=min(f[i+1][Num0],f[i][j]+max(Num0+Num1-j,0));
		}
	}
	int Ans=1e9;
	for(int j=0;j<=30;j++)
		Ans=min(Ans,f[n][j]);
	cout<<Ans<<endl;
	return 0;
}

```