#include <bits/stdc++.h>

using namespace std;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int n, x, a, b;
	vector<int> hist{0};
	
	cin>>n;
	
	while (n--)
	{
		cin>>x;
		a = floor(log2(x)), b = ceil(log2(x));
		
		if (hist.back() <= a)
		{
			hist.push_back(a);
			hist.push_back(b);
		}
		else
		{
			hist.push_back(b);
			hist.push_back(a);
		}
	}
	
	stack<int> s;
	int sol = 0;
	for (int x: hist)
	{
		if (!s.empty() && s.top() >= x)
		{
			sol += s.top() - x;
			
			while (!s.empty() && s.top() >= x)
				s.pop();
		}
		s.push(x);
	}
	sol += s.top();
	
	cout<<sol<<'\n';
}

### Prompt

Your challenge is to write a cpp solution 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>

using namespace std;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int n, x, a, b;
	vector<int> hist{0};
	
	cin>>n;
	
	while (n--)
	{
		cin>>x;
		a = floor(log2(x)), b = ceil(log2(x));
		
		if (hist.back() <= a)
		{
			hist.push_back(a);
			hist.push_back(b);
		}
		else
		{
			hist.push_back(b);
			hist.push_back(a);
		}
	}
	
	stack<int> s;
	int sol = 0;
	for (int x: hist)
	{
		if (!s.empty() && s.top() >= x)
		{
			sol += s.top() - x;
			
			while (!s.empty() && s.top() >= x)
				s.pop();
		}
		s.push(x);
	}
	sol += s.top();
	
	cout<<sol<<'\n';
}

```