#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int n, a[100009], p[100009], q[100009], cnts;

int solve(vector<int>x) {
	if (x.size() == 0) return 0;
	int R = 0, S = 0;
	for (int i = 0; i < x.size(); i++) {
		if (x[i] == 2) S = 1;
		else {
			S++; if (S == 2) { R++; S = 0; }
		}
	}
	if (S >= 1) R++;
	return R;
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i]; int s = a[i], t = 1;
		while (s > 1) { p[i]++; s /= 2; }
		while (t < a[i]) { q[i]++; t *= 2; }
	}
	for (int i = 1; i <= 30; i++) {
		vector<int>P;
		for (int j = 0; j <= n; j++) {
			if (j == n || q[j] < i) {
				cnts += solve(P); P.clear(); continue;
			}
			if (p[j] < i && q[j] >= i) P.push_back(1);
			else P.push_back(2);
		}
	}
	cout << cnts << endl;
	return 0;
}
### Prompt

Create a solution in CPP for 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>
#include <algorithm>
#include <vector>
using namespace std;

int n, a[100009], p[100009], q[100009], cnts;

int solve(vector<int>x) {
	if (x.size() == 0) return 0;
	int R = 0, S = 0;
	for (int i = 0; i < x.size(); i++) {
		if (x[i] == 2) S = 1;
		else {
			S++; if (S == 2) { R++; S = 0; }
		}
	}
	if (S >= 1) R++;
	return R;
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i]; int s = a[i], t = 1;
		while (s > 1) { p[i]++; s /= 2; }
		while (t < a[i]) { q[i]++; t *= 2; }
	}
	for (int i = 1; i <= 30; i++) {
		vector<int>P;
		for (int j = 0; j <= n; j++) {
			if (j == n || q[j] < i) {
				cnts += solve(P); P.clear(); continue;
			}
			if (p[j] < i && q[j] >= i) P.push_back(1);
			else P.push_back(2);
		}
	}
	cout << cnts << endl;
	return 0;
}
```