#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
#define mod 1000000007
int n,a[100005];
vector<int>hoge[35];
bool need[100005];
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		for(int j=0;j<=30;j++){
			if(a[i] <= (1<<j)){
				if(a[i] != (1<<j)) need[i] = 1; 
				a[i] = j; //cout<<j<<endl;
				break;
			}
		}
	}
	int pre = 0; ll ans = 0;
	for(int i=1;i<=n;i++){
		if(pre < a[i]){
			ans+=a[i]-pre; pre = a[i];
		}
		else if(pre > a[i]){
			pre = a[i]; if(need[i]) pre--;
		}
		else if(need[i]){
			pre--;
		}
	}
	cout<<ans<<endl;
}
	
### 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> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
#define mod 1000000007
int n,a[100005];
vector<int>hoge[35];
bool need[100005];
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		for(int j=0;j<=30;j++){
			if(a[i] <= (1<<j)){
				if(a[i] != (1<<j)) need[i] = 1; 
				a[i] = j; //cout<<j<<endl;
				break;
			}
		}
	}
	int pre = 0; ll ans = 0;
	for(int i=1;i<=n;i++){
		if(pre < a[i]){
			ans+=a[i]-pre; pre = a[i];
		}
		else if(pre > a[i]){
			pre = a[i]; if(need[i]) pre--;
		}
		else if(need[i]){
			pre--;
		}
	}
	cout<<ans<<endl;
}
	
```