#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<ctime>
#include<set>
#include<vector>
#include<map>
#include<queue>

#define N 100005
#define M 8000005

#define ls (t<<1)
#define rs ((t<<1)|1)
#define mid ((l+r)>>1)

#define mk make_pair
#define pb push_back
#define fi first
#define se second

using namespace std;

int i,j,m,n,p,k,cnt,a[N],ans;

int main()
{
		scanf("%d",&n);
		for (i=1;i<=n;++i) scanf("%d",&a[i]);
		for (i=29;i>=0;--i)
		{
				int cnt=0;
				for (j=1;j<=n;++j)
					if (a[j]>(1<<i))
				{
					if (a[j]==(1<<(i+1))) ans+=1-cnt,cnt=1;
					else ans+=1-cnt,cnt^=1;
					a[j]=(1<<i);
				}
				else cnt=0;
		}
		printf("%d\n",ans);
}
### Prompt

Please provide a cpp coded solution 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<algorithm>
#include<cstring>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<ctime>
#include<set>
#include<vector>
#include<map>
#include<queue>

#define N 100005
#define M 8000005

#define ls (t<<1)
#define rs ((t<<1)|1)
#define mid ((l+r)>>1)

#define mk make_pair
#define pb push_back
#define fi first
#define se second

using namespace std;

int i,j,m,n,p,k,cnt,a[N],ans;

int main()
{
		scanf("%d",&n);
		for (i=1;i<=n;++i) scanf("%d",&a[i]);
		for (i=29;i>=0;--i)
		{
				int cnt=0;
				for (j=1;j<=n;++j)
					if (a[j]>(1<<i))
				{
					if (a[j]==(1<<(i+1))) ans+=1-cnt,cnt=1;
					else ans+=1-cnt,cnt^=1;
					a[j]=(1<<i);
				}
				else cnt=0;
		}
		printf("%d\n",ans);
}
```