#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int x[200000];
int y[200000];
int dp[200000][2];
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)
	{
		int z;
		scanf("%d", &z);
		for (int j = 0; j < 30; j++)if (z&(1 << j))x[i + 1] = j;
		if ((1 << x[i + 1]) == z)y[i + 1] = 1;
		else x[i + 1]++;
	}
	int cnt = 0;
	dp[0][0] = 1000000000, dp[0][1] = 0;
	for (int i = 1; i <= num; i++)
	{
		if (y[i] == 0)
		{
			dp[i][0] = min(dp[i - 1][0] + max(0, x[i] - x[i - 1] + 1), dp[i - 1][1] + max(0, x[i] - x[i - 1]));
			dp[i][1] = min(dp[i - 1][0] + max(0, x[i] - x[i - 1]), dp[i - 1][1] + max(0, x[i] - x[i - 1] - 1)) + 1;
		}
		else
		{
			dp[i][0] = 1000000000;
			dp[i][1] = min(dp[i - 1][0] + max(0, x[i] - x[i - 1] + 1), dp[i - 1][1] + max(0, x[i] - x[i - 1]));
		}
		//printf("%d %d\n", dp[i][0], dp[i][1]);
	}
	printf("%d\n", min(dp[num][0], dp[num][1]));
}
### Prompt

Please provide a cpp coded solution 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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int x[200000];
int y[200000];
int dp[200000][2];
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)
	{
		int z;
		scanf("%d", &z);
		for (int j = 0; j < 30; j++)if (z&(1 << j))x[i + 1] = j;
		if ((1 << x[i + 1]) == z)y[i + 1] = 1;
		else x[i + 1]++;
	}
	int cnt = 0;
	dp[0][0] = 1000000000, dp[0][1] = 0;
	for (int i = 1; i <= num; i++)
	{
		if (y[i] == 0)
		{
			dp[i][0] = min(dp[i - 1][0] + max(0, x[i] - x[i - 1] + 1), dp[i - 1][1] + max(0, x[i] - x[i - 1]));
			dp[i][1] = min(dp[i - 1][0] + max(0, x[i] - x[i - 1]), dp[i - 1][1] + max(0, x[i] - x[i - 1] - 1)) + 1;
		}
		else
		{
			dp[i][0] = 1000000000;
			dp[i][1] = min(dp[i - 1][0] + max(0, x[i] - x[i - 1] + 1), dp[i - 1][1] + max(0, x[i] - x[i - 1]));
		}
		//printf("%d %d\n", dp[i][0], dp[i][1]);
	}
	printf("%d\n", min(dp[num][0], dp[num][1]));
}
```