#include <iostream>
#define llint long long

using namespace std;

llint n;
llint a[100005];

int main(void)
{
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i];
	
	llint ans = 0, mul = 1;
	bool update = true, flag;
	while(update){
		flag = false, update = false;
		for(int i = 1; i <= n; i++){
			if(a[i] <= mul){
				flag = false;
				continue;
			}
			if(!flag){
				flag = true;
				ans++, update = true;
			}
			else{
				if(a[i] < mul*2) flag = false;
			}
		}
		mul *= 2;
	}
	cout << ans << endl;
	
	return 0;
}
### 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>
#define llint long long

using namespace std;

llint n;
llint a[100005];

int main(void)
{
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i];
	
	llint ans = 0, mul = 1;
	bool update = true, flag;
	while(update){
		flag = false, update = false;
		for(int i = 1; i <= n; i++){
			if(a[i] <= mul){
				flag = false;
				continue;
			}
			if(!flag){
				flag = true;
				ans++, update = true;
			}
			else{
				if(a[i] < mul*2) flag = false;
			}
		}
		mul *= 2;
	}
	cout << ans << endl;
	
	return 0;
}
```