#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <functional>
#include <set>
#include <map>
#include <cmath>
#include <string>
#define SIZE 100005
#define MX 35

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

int A[SIZE];
int dp[2][MX];

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%d",&A[i]);
	int pos=0;
	memset(dp[pos],-1,sizeof(dp[pos]));
	dp[pos][0]=0;
	for(int i=0;i<n;i++)
	{
		int cnt=0,now=1;
		while(now<A[i]) now*=2,cnt++;
		pos^=1;
		memset(dp[pos],-1,sizeof(dp[pos]));
		for(int j=0;j<=cnt;j++)
		{
			if(dp[pos^1][j]==-1) continue;
			int add=cnt-j;
			if(add>0||now==A[i])
			{
				if(dp[pos][cnt]==-1) dp[pos][cnt]=dp[pos^1][j]+add;
				else dp[pos][cnt]=min(dp[pos][cnt],dp[pos^1][j]+add);
			}
			else if(j>0)
			{
				if(dp[pos][cnt-1]==-1) dp[pos][cnt-1]=dp[pos^1][j]+add;
				else dp[pos][cnt-1]=min(dp[pos][cnt-1],dp[pos^1][j]+add);
			}
		}
		for(int j=MX-1;j>0;j--)
		{
			if(dp[pos][j]!=-1)
			{
				if(dp[pos][j-1]==-1) dp[pos][j-1]=dp[pos][j];
				else dp[pos][j-1]=min(dp[pos][j-1],dp[pos][j]);
			}
		}
	}
	printf("%d\n",dp[pos][0]);
	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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <functional>
#include <set>
#include <map>
#include <cmath>
#include <string>
#define SIZE 100005
#define MX 35

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

int A[SIZE];
int dp[2][MX];

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%d",&A[i]);
	int pos=0;
	memset(dp[pos],-1,sizeof(dp[pos]));
	dp[pos][0]=0;
	for(int i=0;i<n;i++)
	{
		int cnt=0,now=1;
		while(now<A[i]) now*=2,cnt++;
		pos^=1;
		memset(dp[pos],-1,sizeof(dp[pos]));
		for(int j=0;j<=cnt;j++)
		{
			if(dp[pos^1][j]==-1) continue;
			int add=cnt-j;
			if(add>0||now==A[i])
			{
				if(dp[pos][cnt]==-1) dp[pos][cnt]=dp[pos^1][j]+add;
				else dp[pos][cnt]=min(dp[pos][cnt],dp[pos^1][j]+add);
			}
			else if(j>0)
			{
				if(dp[pos][cnt-1]==-1) dp[pos][cnt-1]=dp[pos^1][j]+add;
				else dp[pos][cnt-1]=min(dp[pos][cnt-1],dp[pos^1][j]+add);
			}
		}
		for(int j=MX-1;j>0;j--)
		{
			if(dp[pos][j]!=-1)
			{
				if(dp[pos][j-1]==-1) dp[pos][j-1]=dp[pos][j];
				else dp[pos][j-1]=min(dp[pos][j-1],dp[pos][j]);
			}
		}
	}
	printf("%d\n",dp[pos][0]);
	return 0;
}

```