#include <bits/stdc++.h>

using namespace std;

int n, x[100005][2], dp[100005][2], ta;

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &ta);
		for (int j = 0; (1 << j) < ta; j++)
			x[i][1]++;
		if (ta == (1 << x[i][1]))
			x[i][0] = x[i][1];
		else
			x[i][0] = x[i][1] - 1;
		// printf("%d %d\n", x[i][0], x[i][1]);
		dp[i][0] = min(dp[i - 1][0] + max(0, x[i][0] - x[i - 1][1]), dp[i - 1][1] + max(0, x[i][0] - x[i - 1][0])) + x[i][1] - x[i][0];
		dp[i][1] = min(dp[i - 1][0] + max(0, x[i][1] - x[i - 1][1]), dp[i - 1][1] + max(0, x[i][1] - x[i - 1][0]));
	}
	printf("%d\n", min(dp[n][0], dp[n][1]));
	return 0;
}
### Prompt

Please formulate 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 <bits/stdc++.h>

using namespace std;

int n, x[100005][2], dp[100005][2], ta;

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &ta);
		for (int j = 0; (1 << j) < ta; j++)
			x[i][1]++;
		if (ta == (1 << x[i][1]))
			x[i][0] = x[i][1];
		else
			x[i][0] = x[i][1] - 1;
		// printf("%d %d\n", x[i][0], x[i][1]);
		dp[i][0] = min(dp[i - 1][0] + max(0, x[i][0] - x[i - 1][1]), dp[i - 1][1] + max(0, x[i][0] - x[i - 1][0])) + x[i][1] - x[i][0];
		dp[i][1] = min(dp[i - 1][0] + max(0, x[i][1] - x[i - 1][1]), dp[i - 1][1] + max(0, x[i][1] - x[i - 1][0]));
	}
	printf("%d\n", min(dp[n][0], dp[n][1]));
	return 0;
}
```