#include<cstdio>
using namespace std;
const int INF=1e9;
int a[100100];
int dp[100100][32];
int Max(int x,int y){
	return x>y?x:y;
}
int main(){
//	freopen("E.in","r",stdin);
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	int cnt1,cnt;
	for(int i=0;i<=n;i++)
		for(int j=0;j<=31;j++)
			dp[i][j]=INF;
	dp[0][0]=0;
	for(int i=1;i<=n;i++)
	{
		cnt1=cnt=0;
		while(a[i]){
			if(a[i]&1) cnt1++;
			cnt++;
			a[i]>>=1;
		}
		cnt--,cnt1--;
		if(cnt1==0){
			for(int j=0;j<=cnt;j++)
				if(dp[i-1][j]+cnt-j<dp[i][cnt])
					dp[i][cnt]=dp[i-1][j]+cnt-j;
			for(int j=cnt+1;j<=31;j++)
				if(dp[i-1][j]<dp[i][cnt])
					dp[i][cnt]=dp[i-1][j];
		}
		else{
			dp[i][cnt+1]=dp[i-1][0]+cnt+1;
			for(int j=1;j<=cnt;j++)
			{
				if(dp[i-1][j]+cnt-j+1<dp[i][cnt])
					dp[i][cnt]=dp[i-1][j]+cnt-j+1;
				if(dp[i-1][j]+cnt-j+1<dp[i][cnt+1])
					dp[i][cnt+1]=dp[i-1][j]+cnt-j+1;
			}
			for(int j=cnt+1;j<=31;j++)
			{
				if(dp[i-1][j]<dp[i][cnt])
					dp[i][cnt]=dp[i-1][j];
				if(dp[i-1][j]+1<dp[i][cnt+1])
					dp[i][cnt+1]=dp[i-1][j]+1;
			}
		}
	}
	int ans=INF;
	for(int i=0;i<=31;i++)
		if(dp[n][i]<ans)
			ans=dp[n][i];
	printf("%d\n",ans);
	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>
using namespace std;
const int INF=1e9;
int a[100100];
int dp[100100][32];
int Max(int x,int y){
	return x>y?x:y;
}
int main(){
//	freopen("E.in","r",stdin);
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	int cnt1,cnt;
	for(int i=0;i<=n;i++)
		for(int j=0;j<=31;j++)
			dp[i][j]=INF;
	dp[0][0]=0;
	for(int i=1;i<=n;i++)
	{
		cnt1=cnt=0;
		while(a[i]){
			if(a[i]&1) cnt1++;
			cnt++;
			a[i]>>=1;
		}
		cnt--,cnt1--;
		if(cnt1==0){
			for(int j=0;j<=cnt;j++)
				if(dp[i-1][j]+cnt-j<dp[i][cnt])
					dp[i][cnt]=dp[i-1][j]+cnt-j;
			for(int j=cnt+1;j<=31;j++)
				if(dp[i-1][j]<dp[i][cnt])
					dp[i][cnt]=dp[i-1][j];
		}
		else{
			dp[i][cnt+1]=dp[i-1][0]+cnt+1;
			for(int j=1;j<=cnt;j++)
			{
				if(dp[i-1][j]+cnt-j+1<dp[i][cnt])
					dp[i][cnt]=dp[i-1][j]+cnt-j+1;
				if(dp[i-1][j]+cnt-j+1<dp[i][cnt+1])
					dp[i][cnt+1]=dp[i-1][j]+cnt-j+1;
			}
			for(int j=cnt+1;j<=31;j++)
			{
				if(dp[i-1][j]<dp[i][cnt])
					dp[i][cnt]=dp[i-1][j];
				if(dp[i-1][j]+1<dp[i][cnt+1])
					dp[i][cnt+1]=dp[i-1][j]+1;
			}
		}
	}
	int ans=INF;
	for(int i=0;i<=31;i++)
		if(dp[n][i]<ans)
			ans=dp[n][i];
	printf("%d\n",ans);
	return 0;
}

```