#include <cstdio>
int main()
{
	// freopen("CF17-QA-F.in", "r", stdin); 
	int n, cur = 0, ans = 0; 
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		int x; 
		scanf("%d", &x); 
		int p = 0; 
		while (1 << p <= x)
			p++;
		p--;
		bool f = (1 << p) != x; 
		if (cur < p)
			ans += p - cur; 
		else if (cur != p)
			f = false; 
		cur = p; 
		if (f)
		{
			cur++; 
			ans++; 
		}
	}
	printf("%d\n", ans);
	return 0; 
}

### Prompt

Create a solution in Cpp for 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>
int main()
{
	// freopen("CF17-QA-F.in", "r", stdin); 
	int n, cur = 0, ans = 0; 
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		int x; 
		scanf("%d", &x); 
		int p = 0; 
		while (1 << p <= x)
			p++;
		p--;
		bool f = (1 << p) != x; 
		if (cur < p)
			ans += p - cur; 
		else if (cur != p)
			f = false; 
		cur = p; 
		if (f)
		{
			cur++; 
			ans++; 
		}
	}
	printf("%d\n", ans);
	return 0; 
}

```