#include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
ll a[100001];
int b[100001];
int c[100001];

int main(void){
	scanf("%d",&n);
	ll ans=0;
	for(int i=0;i<n;i++){
		scanf("%lld",&a[i]);
		int tmp=0;
		while(a[i]>=(1<<tmp))tmp++;
		tmp--;
		b[i]=tmp;
		if(a[i]>(1<<tmp))c[i]=1;
		ans+=b[i]+c[i];
	}
	int last=0;
	for(int i=0;i<n;i++){
		int m1=min(last,b[i]);
		ans-=m1;
		last-=m1;
		if(last>0 && c[i]>0){
			ans--;
			c[i]--;
		}
		last=b[i]+c[i];
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Your challenge is to write 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>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
ll a[100001];
int b[100001];
int c[100001];

int main(void){
	scanf("%d",&n);
	ll ans=0;
	for(int i=0;i<n;i++){
		scanf("%lld",&a[i]);
		int tmp=0;
		while(a[i]>=(1<<tmp))tmp++;
		tmp--;
		b[i]=tmp;
		if(a[i]>(1<<tmp))c[i]=1;
		ans+=b[i]+c[i];
	}
	int last=0;
	for(int i=0;i<n;i++){
		int m1=min(last,b[i]);
		ans-=m1;
		last-=m1;
		if(last>0 && c[i]>0){
			ans--;
			c[i]--;
		}
		last=b[i]+c[i];
	}
	printf("%lld\n",ans);
	return 0;
}
```