#include<vector>
#include<algorithm>
#include<stdio.h>
using namespace std;
int n;
vector<pair<int,int> > v[30];
int process(int s,int e,int x){
	int sid = lower_bound(v[x].begin(),v[x].end(),make_pair(s,0)) - v[x].begin();
	int eid = upper_bound(v[x].begin(),v[x].end(),make_pair(e,2)) - v[x].begin();
	if(sid < eid){
		int res = 0;
		int P = s;
		int not_payed = 0;
		for(int i=sid; i<eid; i++){
			int Z = v[x][i].first;
			int Q = v[x][i].second;
			if(Z > P){
				res += process(P, Z-1, x+1) + 1;
				not_payed = 0;
			}else
			if(not_payed == 1){
				not_payed = 0;
				res ++;
			}else
			if(Q == 1)
				not_payed = 1;
			P = Z+1;
		}
		if(P <= e){
			res += process(P, e, x+1) + 1;
		}else
		if(not_payed == 1)
			res ++;
		return res;
	}else
		return process(s, e, x+1) + 1;
}
int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; i++){
		int x,t=0;
		scanf("%d",&x);
		while((1<<t) <= x)t++;
		--t;
		v[t].push_back(make_pair(i, (1<<t) == x ? 0 : 1));
	}
	printf("%d\n",process(1,n,0));
	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<vector>
#include<algorithm>
#include<stdio.h>
using namespace std;
int n;
vector<pair<int,int> > v[30];
int process(int s,int e,int x){
	int sid = lower_bound(v[x].begin(),v[x].end(),make_pair(s,0)) - v[x].begin();
	int eid = upper_bound(v[x].begin(),v[x].end(),make_pair(e,2)) - v[x].begin();
	if(sid < eid){
		int res = 0;
		int P = s;
		int not_payed = 0;
		for(int i=sid; i<eid; i++){
			int Z = v[x][i].first;
			int Q = v[x][i].second;
			if(Z > P){
				res += process(P, Z-1, x+1) + 1;
				not_payed = 0;
			}else
			if(not_payed == 1){
				not_payed = 0;
				res ++;
			}else
			if(Q == 1)
				not_payed = 1;
			P = Z+1;
		}
		if(P <= e){
			res += process(P, e, x+1) + 1;
		}else
		if(not_payed == 1)
			res ++;
		return res;
	}else
		return process(s, e, x+1) + 1;
}
int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; i++){
		int x,t=0;
		scanf("%d",&x);
		while((1<<t) <= x)t++;
		--t;
		v[t].push_back(make_pair(i, (1<<t) == x ? 0 : 1));
	}
	printf("%d\n",process(1,n,0));
	return 0;
}
```