#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

int main(){
	LL n;
	cin >> n;
	vector<LL> a;
	a.push_back(1);
	for(LL i = 0; i < n; i++){
		LL c;
		cin >> c;
		a.push_back(c);
	}
	a.push_back(1);
	n += 2;
	LL ans = 0;
	LL last = 0;
	LL up = 0;
	for(LL i = 0; i < n; i++){
		LL c = a[i];
		LL d = a[i]-1;
		LL m1 = -1;
		LL m2 = -1;
		for(LL i = 0; i < 31; i++){
			if(c >= (1LL << i)){
				m1 = i;
			}
			if(d >= (1LL << i)){
				m2 = i;
			}
		}
		m2++;
		if(m1 == m2){
			ans += abs(m1-last);
			last = m1;
			continue;
		} else {
			ans++;
			if(abs(m1-last) < abs(m2-last)){
				ans += abs(m1-last);
				last = m2;
				up = 0;
			} else if(abs(m2-last) < abs(m1-last)){
				ans += abs(m2-last);
				last = m1;
				up = 0;
			} else {
				cout << "HERE" << endl;
			}
		}
	}
	//cout << ans << endl;
	cout << ans/2 << endl;
}

### Prompt

Please create a solution in Cpp 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;
typedef long long LL;

int main(){
	LL n;
	cin >> n;
	vector<LL> a;
	a.push_back(1);
	for(LL i = 0; i < n; i++){
		LL c;
		cin >> c;
		a.push_back(c);
	}
	a.push_back(1);
	n += 2;
	LL ans = 0;
	LL last = 0;
	LL up = 0;
	for(LL i = 0; i < n; i++){
		LL c = a[i];
		LL d = a[i]-1;
		LL m1 = -1;
		LL m2 = -1;
		for(LL i = 0; i < 31; i++){
			if(c >= (1LL << i)){
				m1 = i;
			}
			if(d >= (1LL << i)){
				m2 = i;
			}
		}
		m2++;
		if(m1 == m2){
			ans += abs(m1-last);
			last = m1;
			continue;
		} else {
			ans++;
			if(abs(m1-last) < abs(m2-last)){
				ans += abs(m1-last);
				last = m2;
				up = 0;
			} else if(abs(m2-last) < abs(m1-last)){
				ans += abs(m2-last);
				last = m1;
				up = 0;
			} else {
				cout << "HERE" << endl;
			}
		}
	}
	//cout << ans << endl;
	cout << ans/2 << endl;
}

```