#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int inf = 1012345678;
int main() {
	int N;
	cin >> N;
	vector<int> A(N), D(N);
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
		while ((1 << D[i]) < A[i]) ++D[i];
	}
	int dl = D[0], dr = D[0] + 1;
	if ((1 << D[0]) == A[0]) dr = inf;
	for (int i = 1; i < N; ++i) {
		int nl = min(dl + abs(D[i - 1] - ((1 << D[i]) == A[i] ? D[i] : D[i] - 1)) + ((1 << D[i]) == A[i] ? 0 : 1), dr + abs((D[i - 1] - 1) - ((1 << D[i]) == A[i] ? D[i] : D[i] - 1)) + ((1 << D[i]) == A[i] ? 0 : 1));
		int nr = ((1 << D[i]) == A[i] ? inf : min(dl + abs(D[i - 1] - D[i]) + 1, dr + abs((D[i - 1] - 1) - D[i]) + 1));
		dl = nl;
		dr = nr;
	}
	int ans = min(dl + D[N - 1], dr + D[N - 1] - 1);
	cout << ans / 2 << endl;
	return 0;
}
### Prompt

In cpp, your task is to solve 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 <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int inf = 1012345678;
int main() {
	int N;
	cin >> N;
	vector<int> A(N), D(N);
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
		while ((1 << D[i]) < A[i]) ++D[i];
	}
	int dl = D[0], dr = D[0] + 1;
	if ((1 << D[0]) == A[0]) dr = inf;
	for (int i = 1; i < N; ++i) {
		int nl = min(dl + abs(D[i - 1] - ((1 << D[i]) == A[i] ? D[i] : D[i] - 1)) + ((1 << D[i]) == A[i] ? 0 : 1), dr + abs((D[i - 1] - 1) - ((1 << D[i]) == A[i] ? D[i] : D[i] - 1)) + ((1 << D[i]) == A[i] ? 0 : 1));
		int nr = ((1 << D[i]) == A[i] ? inf : min(dl + abs(D[i - 1] - D[i]) + 1, dr + abs((D[i - 1] - 1) - D[i]) + 1));
		dl = nl;
		dr = nr;
	}
	int ans = min(dl + D[N - 1], dr + D[N - 1] - 1);
	cout << ans / 2 << endl;
	return 0;
}
```