#include <iostream>
using namespace std;

const int maxn = 1e5;
int n, a[maxn], dp[maxn][2], log[maxn];

int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		for (log[i] = 0; 1 << log[i] < a[i]; log[i]++) {}
		if (i == 0)
		{
			dp[i][0] = log[i];
			dp[i][1] = log[i];
		}
		else
		{
			if (a[i] == 1 << log[i])
			{
				dp[i][0] = min(dp[i - 1][0] + max(0, log[i] - (log[i - 1] - 1)), dp[i - 1][1] + max(0, log[i] - log[i - 1]));
				dp[i][1] = dp[i][0];
			}
			else
			{
				dp[i][0] = min(dp[i - 1][0] + max(0, log[i] - (log[i - 1] - 1)), dp[i - 1][1] + max(0, log[i] - log[i - 1]));
				dp[i][1] = 1 + min(dp[i - 1][0] + max(0, (log[i] - 1) - (log[i - 1] - 1)), dp[i - 1][1] + max(0, (log[i] - 1) - log[i - 1]));
			}
		}
	}
	cout << min(dp[n - 1][0], dp[n - 1][1]);
}
### Prompt

Your task is to create 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 <iostream>
using namespace std;

const int maxn = 1e5;
int n, a[maxn], dp[maxn][2], log[maxn];

int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		for (log[i] = 0; 1 << log[i] < a[i]; log[i]++) {}
		if (i == 0)
		{
			dp[i][0] = log[i];
			dp[i][1] = log[i];
		}
		else
		{
			if (a[i] == 1 << log[i])
			{
				dp[i][0] = min(dp[i - 1][0] + max(0, log[i] - (log[i - 1] - 1)), dp[i - 1][1] + max(0, log[i] - log[i - 1]));
				dp[i][1] = dp[i][0];
			}
			else
			{
				dp[i][0] = min(dp[i - 1][0] + max(0, log[i] - (log[i - 1] - 1)), dp[i - 1][1] + max(0, log[i] - log[i - 1]));
				dp[i][1] = 1 + min(dp[i - 1][0] + max(0, (log[i] - 1) - (log[i - 1] - 1)), dp[i - 1][1] + max(0, (log[i] - 1) - log[i - 1]));
			}
		}
	}
	cout << min(dp[n - 1][0], dp[n - 1][1]);
}
```