#include<bits/stdc++.h>
#define B2(x) (63-__builtin_clzll(x))
#define mem(a,b) memset(a,b,sizeof(a))
#define  For(i,a,b) for(int i=a,i##E=b;i<=i##E;++i)
#define rFor(i,a,b) for(int i=a,i##E=b;i>=i##E;--i)
typedef long long LL;
using namespace std;
const int N=100010;
template<typename T>inline bool chkmin(T &a,const T &b){return a>b?a=b,1:0;}
template<typename T>inline bool chkmax(T &a,const T &b){return a<b?a=b,1:0;}
template<typename T>inline void read(T &x)
{
	x=0;int _f(0);char ch=getchar();
	while(!isdigit(ch))_f|=(ch=='-'),ch=getchar();
	while( isdigit(ch))x=x*10+ch-'0',ch=getchar();
	x=_f?-x:x;
}
inline void file()
{
#ifdef ztzshiwo
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
}
int n,ans,l[N],r[N],dp[N][2];
int main()
{
	int x;
	file();
	read(n);
	For(i,1,n)
	{
		read(x);
		l[i]=B2(x);
		r[i]=l[i]+((1<<l[i])!=x);
		//printf("%d: %d l:%d r:%d\n",i,x,l[i],r[i]);
		ans+=r[i];
		dp[i][0]=max(dp[i-1][0]+min(r[i-1],l[i]),dp[i-1][1]+min(l[i-1],l[i]));
		dp[i][1]=max(dp[i-1][0]+min(r[i-1],r[i]),dp[i-1][1]+min(l[i-1],r[i]));
	}
	printf("%d\n",ans-max(dp[n][0],dp[n][1]));
	return 0;
}

### Prompt

Please formulate 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 B2(x) (63-__builtin_clzll(x))
#define mem(a,b) memset(a,b,sizeof(a))
#define  For(i,a,b) for(int i=a,i##E=b;i<=i##E;++i)
#define rFor(i,a,b) for(int i=a,i##E=b;i>=i##E;--i)
typedef long long LL;
using namespace std;
const int N=100010;
template<typename T>inline bool chkmin(T &a,const T &b){return a>b?a=b,1:0;}
template<typename T>inline bool chkmax(T &a,const T &b){return a<b?a=b,1:0;}
template<typename T>inline void read(T &x)
{
	x=0;int _f(0);char ch=getchar();
	while(!isdigit(ch))_f|=(ch=='-'),ch=getchar();
	while( isdigit(ch))x=x*10+ch-'0',ch=getchar();
	x=_f?-x:x;
}
inline void file()
{
#ifdef ztzshiwo
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
}
int n,ans,l[N],r[N],dp[N][2];
int main()
{
	int x;
	file();
	read(n);
	For(i,1,n)
	{
		read(x);
		l[i]=B2(x);
		r[i]=l[i]+((1<<l[i])!=x);
		//printf("%d: %d l:%d r:%d\n",i,x,l[i],r[i]);
		ans+=r[i];
		dp[i][0]=max(dp[i-1][0]+min(r[i-1],l[i]),dp[i-1][1]+min(l[i-1],l[i]));
		dp[i][1]=max(dp[i-1][0]+min(r[i-1],r[i]),dp[i-1][1]+min(l[i-1],r[i]));
	}
	printf("%d\n",ans-max(dp[n][0],dp[n][1]));
	return 0;
}

```