//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
const int N=1e5+5;
int a[N],b[N],c[N],n,res,lst,tmp;
int main(){
	scanf("%d",&n);
	for(R int i=1,x;i<=n;++i){
		scanf("%d",&x),a[i]=x,b[i]=-1;
		while(a[i])++b[i],a[i]>>=1;
		c[i]=((x&-x)!=x),res+=b[i]+c[i];
	}
	fp(i,1,n){
		tmp=min(lst,b[i]),res-=tmp,lst-=tmp;
		if(lst&&c[i])--res,--c[i];
		lst=b[i]+c[i];
	}
	printf("%d\n",res);
	return 0;
}
### 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
//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
const int N=1e5+5;
int a[N],b[N],c[N],n,res,lst,tmp;
int main(){
	scanf("%d",&n);
	for(R int i=1,x;i<=n;++i){
		scanf("%d",&x),a[i]=x,b[i]=-1;
		while(a[i])++b[i],a[i]>>=1;
		c[i]=((x&-x)!=x),res+=b[i]+c[i];
	}
	fp(i,1,n){
		tmp=min(lst,b[i]),res-=tmp,lst-=tmp;
		if(lst&&c[i])--res,--c[i];
		lst=b[i]+c[i];
	}
	printf("%d\n",res);
	return 0;
}
```