#include <bits/stdc++.h>
#define FOR(i,a,b) for(ut i=(a);i<(ut)(b);i++)
#define REP(i,b) FOR(i,0,b)
#define ALL(c) c.begin(),c.end()
#define PB push_back
#define cat //cout << __LINE__ << endl;
using namespace std;
typedef long long LL;
typedef double ld;
typedef LL ut;
typedef vector<ut> VI;
typedef pair<ut,ut> pr;
typedef pair<ut,pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
typedef priority_queue<ut,VI> PQ;
const int SIZE=2*1e5;
const LL INF=1<<30;
const LL p=7+1e9;
LL A[SIZE];
LL B[SIZE],C[SIZE];
int main(){
	LL N;
	cin >> N;
	REP(i,N){
		cin >> A[i];
		B[i]=log2(A[i]);
		C[i]=A[i]!=(1LL<<B[i]);
	}
	LL ans=0;
	while(true){
		bool exist=false;
		bool now=false;
		REP(i,N){
			if(!now && B[i]+C[i]>=1){
				if(C[i]>=1) C[i]--;
				else B[i]--;
				now=true;
				ans++;
				exist=true;
			}
			else if(now && B[i]){
				B[i]--;
			}
			else if(now && C[i]){
				C[i]--;
				now=false;
			}else{
				now=false;
			}
		}
		if(!exist) break;
	}
	cout << ans << endl;

	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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 FOR(i,a,b) for(ut i=(a);i<(ut)(b);i++)
#define REP(i,b) FOR(i,0,b)
#define ALL(c) c.begin(),c.end()
#define PB push_back
#define cat //cout << __LINE__ << endl;
using namespace std;
typedef long long LL;
typedef double ld;
typedef LL ut;
typedef vector<ut> VI;
typedef pair<ut,ut> pr;
typedef pair<ut,pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
typedef priority_queue<ut,VI> PQ;
const int SIZE=2*1e5;
const LL INF=1<<30;
const LL p=7+1e9;
LL A[SIZE];
LL B[SIZE],C[SIZE];
int main(){
	LL N;
	cin >> N;
	REP(i,N){
		cin >> A[i];
		B[i]=log2(A[i]);
		C[i]=A[i]!=(1LL<<B[i]);
	}
	LL ans=0;
	while(true){
		bool exist=false;
		bool now=false;
		REP(i,N){
			if(!now && B[i]+C[i]>=1){
				if(C[i]>=1) C[i]--;
				else B[i]--;
				now=true;
				ans++;
				exist=true;
			}
			else if(now && B[i]){
				B[i]--;
			}
			else if(now && C[i]){
				C[i]--;
				now=false;
			}else{
				now=false;
			}
		}
		if(!exist) break;
	}
	cout << ans << endl;

	return 0;
}
```