Problem: You are playing a game and your goal is to maximize your expected gain. At the beginning of the game, a pawn is put, uniformly at random, at a position p\in\\{1,2,\dots, N\\}. The N positions are arranged on a circle (so that 1 is between N and 2).

The game consists of turns. At each turn you can either end the game, and get A_p dollars (where p is the current position of the pawn), or pay B_p dollar to keep playing. If you decide to keep playing, the pawn is randomly moved to one of the two adjacent positions p-1, p+1 (with the identifications 0 = N and N+1=1).

What is the expected gain of an optimal strategy?

Note: The "expected gain of an optimal strategy" shall be defined as the supremum of the expected gain among all strategies such that the game ends in a finite number of turns.

Constraints

* 2 \le N \le 200,000
* 0 \le A_p \le 10^{12} for any p = 1,\ldots, N
* 0 \le B_p \le 100 for any p = 1, \ldots, N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N


Output

Print a single real number, the expected gain of an optimal strategy. Your answer will be considered correct if its relative or absolute error does not exceed 10^{-10}.

Examples

Input

5
4 2 6 3 5
1 1 1 1 1


Output

4.700000000000


Input

4
100 0 100 0
0 100 0 100


Output

50.000000000000


Input

14
4839 5400 6231 5800 6001 5200 6350 7133 7986 8012 7537 7013 6477 5912
34 54 61 32 52 61 21 43 65 12 45 21 1 4


Output

7047.142857142857


Input

10
470606482521 533212137322 116718867454 746976621474 457112271419 815899162072 641324977314 88281100571 9231169966 455007126951
26 83 30 59 100 88 84 91 54 61


Output

815899161079.400024414062

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<ctime>
#include<map>
#include<bitset>
#include<set>
#include<assert.h>
#include<chrono>
#include<random>
#include<iostream>
#define LL long long
#define mp(x,y) make_pair(x,y)
#define pll pair<long long,long long>
#define pii pair<int,int>
#define SZ(x) ((int)x.size())
#define VI vector<int>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
inline LL read()
{
	LL f=1,x=0;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int stack[20];
template<typename T>inline void write(T x)
{
	if(x<0){putchar('-');x=-x;}
    if(!x){putchar('0');return;}
    int top=0;
    while(x)stack[++top]=x%10,x/=10;
    while(top)putchar(stack[top--]+'0');
}
template<typename T>inline void pr1(T x){write(x);putchar(' ');}
template<typename T>inline void pr2(T x){write(x);putchar('\n');}
template<typename T>inline void chkmin(T &x,T y){x=x<y?x:y;}
template<typename T>inline void chkmax(T &x,T y){x=x>y?x:y;}
const int MAXN=200005;
struct pnt
{
	double x,y;pnt(){}pnt(double _x,double _y){x=_x;y=_y;}
	friend pnt operator +(pnt n1,pnt n2){return pnt(n1.x+n2.x,n1.y+n2.y);}
	friend pnt operator -(pnt n1,pnt n2){return pnt(n1.x-n2.x,n1.y-n2.y);}
	friend double operator *(pnt n1,pnt n2){return n1.x*n2.y-n2.x*n1.y;}
}sta[MAXN];int tp;
double A[MAXN],B[MAXN],C[MAXN];
double newA[MAXN],newB[MAXN];
int n;
void Shift()
{
	int p=0;
	for(int i=1;i<=n;i++)if(A[i]>A[p])p=i;int t=0;
	for(int i=p+1;i<=n;i++)newA[++t]=A[i];
	for(int i=1;i<p;i++)newA[++t]=A[i];t=0;
	for(int i=p+1;i<=n;i++)newB[++t]=B[i];
	for(int i=1;i<p;i++)newB[++t]=B[i];
	A[0]=A[n]=A[p];B[0]=B[n]=B[p];
	for(int i=1;i<n;i++)A[i]=newA[i],B[i]=newB[i];
}
int main()
{
	#ifdef Rose
		double BeginJudgeTime=clock();
	#endif
	n=read();
	for(int i=1;i<=n;i++)A[i]=read();
	for(int i=1;i<=n;i++)B[i]=read();Shift();
	if(n-1==1)C[1]=B[1];
	else for(int i=2;i<=n;i++)C[i]=2*C[i-1]-C[i-2]-2*B[i-1];
	sta[1]=pnt(0,A[0]+C[0]);sta[2]=pnt(1,A[1]+C[1]);tp=2;
	for(int i=2;i<=n;i++)
	{
		pnt now=pnt(i,A[i]+C[i]);
		while(tp>=2&&(now-sta[tp-1])*(sta[tp]-sta[tp-1])<=0)--tp;
		sta[++tp]=now;
	}double sum=0;
	for(int i=1,p=1;i<n;i++)
	{
		while(p+1<tp&&sta[p+1].x<=i)++p;
		double ans=sta[p].y+(sta[p+1].y-sta[p].y)*((i-sta[p].x)/(sta[p+1].x-sta[p].x));
		ans-=C[i];
		sum+=ans/(1.0*n);
	}
	sum+=A[0]/(1.0*n);
	printf("%.10lf\n",sum);
	#ifdef Rose
		double EndJudgeTime=clock();
		cerr<<"JudgeTime is"<<" ";
		cerr<<(EndJudgeTime-BeginJudgeTime)/CLOCKS_PER_SEC<<endl;
	#endif
	return 0;
}