#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const int MC = 100010;
int N;
int a[MC];
int ans;

int main(){
	cin >> N;
	repp(i,0,N) cin >> a[i];
	int z = 1;
	while(z*2<(int)(1e9)) z <<= 1;
	while(z>0){
		bool b = 0;
		repp(i,0,N){
			if(a[i] > z){
				if(a[i] < z*2){
					if(b){
						++ans;
					}
					b = !b;
				} else {
					b = 1;
				}
				a[i] = z;
			} else if(b){
				++ans;
				b = 0;
			}
		}
		if(b) ++ans;
		z >>= 1;
	}
	cout << ans << 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;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const int MC = 100010;
int N;
int a[MC];
int ans;

int main(){
	cin >> N;
	repp(i,0,N) cin >> a[i];
	int z = 1;
	while(z*2<(int)(1e9)) z <<= 1;
	while(z>0){
		bool b = 0;
		repp(i,0,N){
			if(a[i] > z){
				if(a[i] < z*2){
					if(b){
						++ans;
					}
					b = !b;
				} else {
					b = 1;
				}
				a[i] = z;
			} else if(b){
				++ans;
				b = 0;
			}
		}
		if(b) ++ans;
		z >>= 1;
	}
	cout << ans << endl;
	return 0;
}

```