#include <bits/stdc++.h>
using namespace std;
int main(){
	int n; cin >> n;
	using ii = pair<int,int>;
	vector<ii> a(n);
	int ans = 0;
	for(int i = 0; i < n; i++){
		int x; cin >> x;
		int k = 0;
		while((1<<k) <= x) k++;
		k--;
		if((1<<k) == x){
			a[i] = {k, k};
			ans += k;
		} else {
			a[i] = {k, k+1};
			ans += k + 1;
		}
	}
	vector<int> f(2, 0), nf(2, 0);
	for(int i = 1; i < n; i++){
		int p, q, r, s;
		tie(p, q) = a[i - 1];
		tie(r, s) = a[i];
		nf[0] = max(f[0] + min(p, s), f[1] + min(q, s));
		nf[1] = max(f[0] + min(p, r), f[1] + min(q, r));
		f = nf;
	}
	cout << ans - max(f[0], f[1]) << endl;
	return 0;
}

### Prompt

Generate 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; cin >> n;
	using ii = pair<int,int>;
	vector<ii> a(n);
	int ans = 0;
	for(int i = 0; i < n; i++){
		int x; cin >> x;
		int k = 0;
		while((1<<k) <= x) k++;
		k--;
		if((1<<k) == x){
			a[i] = {k, k};
			ans += k;
		} else {
			a[i] = {k, k+1};
			ans += k + 1;
		}
	}
	vector<int> f(2, 0), nf(2, 0);
	for(int i = 1; i < n; i++){
		int p, q, r, s;
		tie(p, q) = a[i - 1];
		tie(r, s) = a[i];
		nf[0] = max(f[0] + min(p, s), f[1] + min(q, s));
		nf[1] = max(f[0] + min(p, r), f[1] + min(q, r));
		f = nf;
	}
	cout << ans - max(f[0], f[1]) << endl;
	return 0;
}

```