#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x)
{
    x=0; char c=getchar(); int f=1;
    while(!isdigit(c)){if(c=='-')f=-1; c=getchar();}
    while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=1e5+10;
int a[N],n,c[N][2];
int Log(int x)
{
//	cout<<x<<' ';
	int cnt=0; x--;
	while(x) cnt++,x>>=1;
//	cout<<cnt<<endl;
	return cnt;
}
int dp[N][2];
int main()
{
	rd(n); ll ans=0;
	for(int i=1;i<=n;++i) rd(a[i]),c[i][0]=Log(a[i]),c[i][1]=((1<<c[i][0])==a[i])?c[i][0]:c[i][0]-1,ans+=c[i][0];
	for(int i=1;i<=n;++i)
		for(int j=0;j<2;++j)
			for(int k=0;k<2;++k)
				dp[i][j]=max(dp[i][j],dp[i-1][k]+min(c[i-1][k^1],c[i][j]));
//	cout<<endl;
	printf("%lld\n",ans-max(dp[n][0],dp[n][1]));
	return 0;
}
### 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x)
{
    x=0; char c=getchar(); int f=1;
    while(!isdigit(c)){if(c=='-')f=-1; c=getchar();}
    while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=1e5+10;
int a[N],n,c[N][2];
int Log(int x)
{
//	cout<<x<<' ';
	int cnt=0; x--;
	while(x) cnt++,x>>=1;
//	cout<<cnt<<endl;
	return cnt;
}
int dp[N][2];
int main()
{
	rd(n); ll ans=0;
	for(int i=1;i<=n;++i) rd(a[i]),c[i][0]=Log(a[i]),c[i][1]=((1<<c[i][0])==a[i])?c[i][0]:c[i][0]-1,ans+=c[i][0];
	for(int i=1;i<=n;++i)
		for(int j=0;j<2;++j)
			for(int k=0;k<2;++k)
				dp[i][j]=max(dp[i][j],dp[i-1][k]+min(c[i-1][k^1],c[i][j]));
//	cout<<endl;
	printf("%lld\n",ans-max(dp[n][0],dp[n][1]));
	return 0;
}
```