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

### Prompt

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

```