#include <bits/stdc++.h>
using namespace std;
#define N 2005
#define ll long long
struct Vec
{
	int x,y;
	Vec(int a=0,int b=0) {x=a,y=b;}
	double norm() {return sqrt(1LL*x*x+1LL*y*y);}
};
const double inf=1e100;
Vec operator + (const Vec &x,const Vec &y) {return Vec(x.x+y.x,x.y+y.y);}
Vec operator - (const Vec &x,const Vec &y) {return Vec(x.x-y.x,x.y-y.y);}
ll cross(const Vec &x,const Vec &y) {return 1LL*x.x*y.y-1LL*x.y*y.x;}
int sgn(ll x)
{
	if(x<0) return -1;
	else if(x>0) return 1;
	else return 0;
}
int ccw(const Vec &x,const Vec &y,const Vec &z) {return sgn(cross(y-x,z-x));}
Vec a[N];
Vec o(0,0);
int id[N];
int n,ok[N][N];
double dis[N][N],ans[N][N];
signed main() {
#ifdef xay5421
	freopen("b.in","r",stdin);
#endif
	cin>>n;
	for(int i=0;i<n;i++) cin>>a[i].x>>a[i].y;
	for(int i=0;i<n;i++) id[i]=i;
	sort(id,id+n,[&](int x,int y){
		return atan2(a[x].y,a[x].x)<atan2(a[y].y,a[y].x);
	});
	// for(int i=0;i<n;i++) printf("%d %d\n",a[i].x,a[i].y);
	for(int i=0;i<n;i++)
	{
		Vec premn=a[id[i]],u=a[id[i]];
		for(int j=1;j<n;j++)
		{
			Vec v=a[id[(i+j)%n]];
			if(ccw(o,u,v)==-1) break;
			if(ccw(u,premn,v)==-1) {}
			else ok[id[i]][id[(i+j)%n]]=1,premn=v;
		}
		premn=u;
		for(int j=n-1;j>=1;j--)
		{
			Vec v=a[id[(i+j)%n]];
			if(ccw(o,u,v)==1) break;
			if(ccw(u,premn,v)==1) {}
			else ok[id[i]][id[(i+j)%n]]=1,premn=v;
		}
	}
	// for(int i=0;i<n;i++) for(int j=0;j<n;j++) printf("%d%c",ok[i][j]," \n"[j==n-1]);
	for(int i=0;i<n;i++) for(int j=0;j<n;j++)
	{
		if(i==j) dis[i][j]=0;
		else if(ok[i][j]) dis[i][j]=(a[i]-a[j]).norm();
		else dis[i][j]=inf;
	}
	cerr<<clock()<<endl;
	for(int k=0;k<n;k++) for(int i=0;i<n;i++) for(int j=0;j<n;j++) dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
	cerr<<clock()<<endl;
	for(int i=0;i<n;i++) for(int j=i;j<n;j++)
	{
		for(int k=0;k<n;k++) ans[i+1][j+1]=max(ans[i+1][j+1],dis[i][k]+dis[j][k]);
		ans[j+1][i+1]=ans[i+1][j+1];
	}
	cerr<<clock()<<endl;
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) ans[i][j]=ans[i][j]+ans[i-1][j]+ans[i][j-1]-ans[i-1][j-1];
	int Q; cin>>Q;
	while(Q--)
	{
		int l1,r1,l2,r2;
		scanf("%d %d %d %d",&l1,&r1,&l2,&r2);
		double sum=ans[r1][r2]-ans[l1-1][r2]-ans[r1][l2-1]+ans[l1-1][l2-1];
		printf("%.10lf\n",sum/(r1-l1+1)/(r2-l2+1));
	}
	
	return 0;
}