#include <bits/stdc++.h>
using namespace std;
int main() {
	int n, x;
	scanf("%d", &n);
	scanf("%d", &x);
	int y = ilogb(x);
	vector<pair<long long, int>> all;
	if ((1 << y) == x) all.push_back({y, y}); else all.push_back({y + 1LL, y + 1});
	for (int i = 2; i <= n; ++ i) {
		scanf("%d", &x);
		y = ilogb(x);
		vector<int> l, r, m;
		vector<pair<long long, int>> bll;
		if ((1 << y) == x) {
			l.push_back(y);
			r.push_back(y);
			m.push_back(y);
		}
		else {
			l.push_back(y);
			r.push_back(y + 1);
			m.push_back(y + 1);
			l.push_back(y + 1);
			r.push_back(y);
			m.push_back(y + 1);
		}
		for (int i = 0; i < (int) l.size(); ++ i) {
			long long res = 1e18;
			for (auto p : all) {
				long long tot = p.first + m[i] - min(l[i], p.second);
				res = min(res, tot);
			}
			bll.push_back({res, r[i]});
		}
		all.swap(bll);
	}
	long long ans = 1e18;
	for (auto p : all) ans = min(ans, p.first);
	printf("%lld\n", ans);
}


### 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 main() {
	int n, x;
	scanf("%d", &n);
	scanf("%d", &x);
	int y = ilogb(x);
	vector<pair<long long, int>> all;
	if ((1 << y) == x) all.push_back({y, y}); else all.push_back({y + 1LL, y + 1});
	for (int i = 2; i <= n; ++ i) {
		scanf("%d", &x);
		y = ilogb(x);
		vector<int> l, r, m;
		vector<pair<long long, int>> bll;
		if ((1 << y) == x) {
			l.push_back(y);
			r.push_back(y);
			m.push_back(y);
		}
		else {
			l.push_back(y);
			r.push_back(y + 1);
			m.push_back(y + 1);
			l.push_back(y + 1);
			r.push_back(y);
			m.push_back(y + 1);
		}
		for (int i = 0; i < (int) l.size(); ++ i) {
			long long res = 1e18;
			for (auto p : all) {
				long long tot = p.first + m[i] - min(l[i], p.second);
				res = min(res, tot);
			}
			bll.push_back({res, r[i]});
		}
		all.swap(bll);
	}
	long long ans = 1e18;
	for (auto p : all) ans = min(ans, p.first);
	printf("%lld\n", ans);
}


```