#include<cstdio>
#include<algorithm>
#include<vector>
#include<deque>
#include<stack>
#include<queue>
#include<string>
#include<iostream>
#include<tuple>
#include<utility>
#include<set>
#include<queue>
#include<iomanip>
#include<iterator>
#include<map>
//#include<chrono>
//cout<<setprecision(12)
//fixed
//#include<random>
using namespace std;
typedef long long int llint;
typedef long double lldo;
const llint big=1e9+100000;
const int mod=1e9+7;
const lldo eps=1e-9;
const long double pai=3.141592653589793238462643;
#define mt make_tuple
#define mp make_pair
#define fir first
#define sec second
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define res resize
#define ins insert
#define era erase
#define dme(in) cout<<in<<endl;return 0
template <class T,class U>void mineq(T& a,U b){if(a>b){a=b;}}
template <class T,class U>void maxeq(T& a,U b){if(a<b){a=b;}}
int main(void){
	int n;cin>>n;
	llint st=0,ue=1,mae=0;
	//st maeそのものの数
	//ue mae+1回合成する
	for(int i=0;i<=n;i++){
		llint q;
		if(i!=n){cin>>q;}else{q=1;}
		llint bai=0;
		for(bai=0;(1LL<<bai)<q;bai++){}
		if((1LL<<bai)==q){
			//cout<<"da"<<bai<<endl;
			st=min(st+abs(mae-bai),ue+abs(mae+1-bai));
			ue=st+1;
		}else{
			bai--;
			//cout<<"de"<<bai<<endl;
			llint nst=min(st+abs(bai-mae),ue+abs(bai-mae-1));
			llint nue=min(st+abs(bai+1-mae),ue+abs(bai+1-mae-1));
			ue=nst+1;
			st=nue+1;
		}
		mae=bai;
	}
	cout<<st/2<<endl;
	return 0;
}
### 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<cstdio>
#include<algorithm>
#include<vector>
#include<deque>
#include<stack>
#include<queue>
#include<string>
#include<iostream>
#include<tuple>
#include<utility>
#include<set>
#include<queue>
#include<iomanip>
#include<iterator>
#include<map>
//#include<chrono>
//cout<<setprecision(12)
//fixed
//#include<random>
using namespace std;
typedef long long int llint;
typedef long double lldo;
const llint big=1e9+100000;
const int mod=1e9+7;
const lldo eps=1e-9;
const long double pai=3.141592653589793238462643;
#define mt make_tuple
#define mp make_pair
#define fir first
#define sec second
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define res resize
#define ins insert
#define era erase
#define dme(in) cout<<in<<endl;return 0
template <class T,class U>void mineq(T& a,U b){if(a>b){a=b;}}
template <class T,class U>void maxeq(T& a,U b){if(a<b){a=b;}}
int main(void){
	int n;cin>>n;
	llint st=0,ue=1,mae=0;
	//st maeそのものの数
	//ue mae+1回合成する
	for(int i=0;i<=n;i++){
		llint q;
		if(i!=n){cin>>q;}else{q=1;}
		llint bai=0;
		for(bai=0;(1LL<<bai)<q;bai++){}
		if((1LL<<bai)==q){
			//cout<<"da"<<bai<<endl;
			st=min(st+abs(mae-bai),ue+abs(mae+1-bai));
			ue=st+1;
		}else{
			bai--;
			//cout<<"de"<<bai<<endl;
			llint nst=min(st+abs(bai-mae),ue+abs(bai-mae-1));
			llint nue=min(st+abs(bai+1-mae),ue+abs(bai+1-mae-1));
			ue=nst+1;
			st=nue+1;
		}
		mae=bai;
	}
	cout<<st/2<<endl;
	return 0;
}
```