#include <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<55;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

int n;
vi a;

int main(){
	cin>>n;
	a=vi(n);
	for(int i=0;i<n;i++) cin>>a[i];
	int res=0,t=0;
	for(int i=0;i<n;i++){
		int t_=0;
		while(a[i]>1){
			if(a[i]%2==1){
				int I=-1;
				for(int j=30;j>=0;j--) if(a[i]&1<<j){
					if(I!=-1) a[i]-=1<<j;
					else I=j;
				}
				if(t>I) t--;
				else t_++,res++;
			}
			else{
				a[i]/=2;
				if(t) t--;
				else res++;
				t_++;
			}
		}
		t=t_;
	}
	cout<<res<<endl;
}
### Prompt

In cpp, your task is to solve 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>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<55;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

int n;
vi a;

int main(){
	cin>>n;
	a=vi(n);
	for(int i=0;i<n;i++) cin>>a[i];
	int res=0,t=0;
	for(int i=0;i<n;i++){
		int t_=0;
		while(a[i]>1){
			if(a[i]%2==1){
				int I=-1;
				for(int j=30;j>=0;j--) if(a[i]&1<<j){
					if(I!=-1) a[i]-=1<<j;
					else I=j;
				}
				if(t>I) t--;
				else t_++,res++;
			}
			else{
				a[i]/=2;
				if(t) t--;
				else res++;
				t_++;
			}
		}
		t=t_;
	}
	cout<<res<<endl;
}
```