#include <bits/stdc++.h>
#define Maxn 100007
using namespace std;
int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,ans=0;
int a[Maxn],l,r,q=0;
int main()
{
	n=read();
	for (int i=1;i<=n;i++)
		a[i]=read();
	for (int i=1;i<=n;i++)
	{
		int now=a[i];
		l=0;
		while (now>1)
		{
			++l;
			now/=2;
		}
		if (a[i]!=(1<<l)) r=1; else r=0;
		if (l>q) ans+=l-q; if (l<q) r=0;
		q=l;
		if (r==1)
		{
			++ans;
			++q;
		}
	}
	printf("%d\n",ans);
	return 0;
}

### Prompt

Please create a solution in Cpp to 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 <bits/stdc++.h>
#define Maxn 100007
using namespace std;
int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,ans=0;
int a[Maxn],l,r,q=0;
int main()
{
	n=read();
	for (int i=1;i<=n;i++)
		a[i]=read();
	for (int i=1;i<=n;i++)
	{
		int now=a[i];
		l=0;
		while (now>1)
		{
			++l;
			now/=2;
		}
		if (a[i]!=(1<<l)) r=1; else r=0;
		if (l>q) ans+=l-q; if (l<q) r=0;
		q=l;
		if (r==1)
		{
			++ans;
			++q;
		}
	}
	printf("%d\n",ans);
	return 0;
}

```