#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int A[1100000];
int n,dp[110000][40];
const int M=35;
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&A[i]);
	int pre=0;
	memset(dp,0x3f,sizeof dp); dp[0][0]=0;
	for (int i=1;i<=n;i++){
		int num=0,flag=0;
		int x=A[i];
		while (x>1){
			x=(x+1)/2; num++;
		}
		x=A[i];
		while (x>1){
			if (x&1) flag=1; x>>=1;
		}
		if (flag==0){
			for (int j=0;j<=M;j++)
				dp[i][num]=min(dp[i][num],dp[i-1][j]+max(0,num-j));
		} else {
			for (int j=1;j<=M;j++)
				dp[i][num-1]=min(dp[i][num-1],dp[i-1][j]+max(0,num-j));
			for (int j=0;j<=M;j++)
				dp[i][num]=min(dp[i][num],dp[i-1][j]+max(0,num-1-j)+1);
		}
	}
	int ans=1e9;
	for (int i=0;i<=M;i++) ans=min(ans,dp[n][i]);
	cout<<ans<<endl;
}
### Prompt

In CPP, your task is to solve 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<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int A[1100000];
int n,dp[110000][40];
const int M=35;
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&A[i]);
	int pre=0;
	memset(dp,0x3f,sizeof dp); dp[0][0]=0;
	for (int i=1;i<=n;i++){
		int num=0,flag=0;
		int x=A[i];
		while (x>1){
			x=(x+1)/2; num++;
		}
		x=A[i];
		while (x>1){
			if (x&1) flag=1; x>>=1;
		}
		if (flag==0){
			for (int j=0;j<=M;j++)
				dp[i][num]=min(dp[i][num],dp[i-1][j]+max(0,num-j));
		} else {
			for (int j=1;j<=M;j++)
				dp[i][num-1]=min(dp[i][num-1],dp[i-1][j]+max(0,num-j));
			for (int j=0;j<=M;j++)
				dp[i][num]=min(dp[i][num],dp[i-1][j]+max(0,num-1-j)+1);
		}
	}
	int ans=1e9;
	for (int i=0;i<=M;i++) ans=min(ans,dp[n][i]);
	cout<<ans<<endl;
}
```