Problem: Problem

Given a convex polygon consisting of $ N $ vertices and the center coordinates of $ M $ circles. The radius of all circles is $ r $.

I want to find the minimum real number $ r $ that satisfies the following conditions.

Condition: Every point inside a convex polygon is contained in at least one circle.

Constraints

The input satisfies the following conditions.

* $ 3 \ le N \ le 100 $
* $ 1 \ le M \ le 100 $
* $ -10 ^ 5 \ le px_i, py_i \ le 10 ^ 5 $
* $ -10 ^ 5 \ le cx_i, cy_i \ le 10 ^ 5 $
* No matter which of the three vertices of the convex polygon is selected, they do not exist on the same straight line.

Input

All inputs are given as integers in the following format:


$ N $
$ px_1 $ $ py_1 $
$ px_2 $ $ py_2 $
::
$ px_N $ $ py_N $
$ M $
$ Cx_1 $ $ cy_1 $
$ cx_2 $ $ cy_2 $
::
$ Cx_M $ $ cy_M $


The $ 1 $ line is given an integer $ N $ that represents the number of vertices of the convex polygon.
Information on each vertex of the convex polygon is given in the counterclockwise order from the $ 2 $ line to the $ 2 + N-1 $ line. The $ 1 + i $ line is given $ px_i $ $ py_i $, which represents the coordinates of the $ i $ th vertex, separated by blanks.
The $ 2 + N $ line is given the integer $ M $, which represents the number of circles.
Information on the center coordinates of the circle is given from the $ 2 + N + 1 $ line to the $ 2 + N + M $ line. On the $ 2 + N + i $ line, $ cx_i $ $ cy_i $ representing the center coordinates of the $ i $ th circle is given, separated by blanks.

Output

Outputs the radius $ r $ of the smallest circle that satisfies the condition.
However, absolute or relative errors up to $ 10 ^ {-5} $ are acceptable.

Examples

Input

4
2 3
1 2
2 1
3 2
4
1 3
1 1
3 3
3 1


Output

1.414213562373


Input

7
-96469 25422
-55204 -45592
-29140 -72981
98837 -86795
92303 63297
19059 96012
-67980 70342
17
-4265 -14331
33561 72343
52055 98952
-71189 60872
10459 -7512
-11981 57756
-78228 -28044
37397 -69980
-27527 -51966
22661 -16694
13759 -59976
86539 -47703
17098 31709
-62497 -70998
-57608 59799
-1904 -35574
-73860 121


Output

75307.220122044484

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 1000000000000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;




#define SIZE 105

struct Point{
	Point(double arg_x,double arg_y){
		x = arg_x;
		y = arg_y;
	}

	Point(){
		x = y = 0.0;
	}

	Point operator + (Point p){ return Point(x+p.x,y+p.y); }
	Point operator - (Point p){ return Point(x-p.x,y-p.y);}
	Point operator * (double a){ return Point(a*x,a*y); }
	Point operator / (double a){ return Point(x/a,y/a); }

	double abs(){ return sqrt(norm()); }
	double norm(){ return x*x + y*y; }

	bool operator<(const Point &p) const{
		return x != p.x? x < p.x: y < p.y;
	}

	bool operator == (const Point &p) const{
		return fabs(x-p.x) < EPS && fabs(y-p.y) < EPS;
	}

	double x,y;
};

typedef Point Vector;
typedef vector<Point> Polygon;

struct Line{
	Line(){

	}
	Line(Point a,Point b){
		p[0] = a;
		p[1] = b;
	}
	/*void outPut(){

		printf("(%.3lf,%.3lf)-(%.3lf,%.3lf)\n",p[0].x,p[0].y,p[1].x,p[1].y);
	}*/
	Point p[2];
};

int N,num_circle;
double NUM = 20005;
double E[SIZE];

Point C[SIZE];
Polygon L;


double cross(Vector a,Vector b){
    return a.x*b.y-a.y*b.x;
}

double dot(Vector a,Vector b){
    return a.x*b.x + a.y*b.y;
}

static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -1;
static const int ON_SEGMENT = 0;

int ccw(Point p0,Point p1,Point p2){

	Vector a = p1 - p0;
	Vector b = p2 - p0;

	if(cross(a,b) > EPS)return COUNTER_CLOCKWISE;
	if(cross(a,b) < -EPS)return CLOCKWISE;
	if(dot(a,b) < -EPS)return ONLINE_BACK;
	if(a.norm() < b.norm())return ONLINE_FRONT;

	return ON_SEGMENT;
}

Point calc_Cross_Point(double x1,double x2,double x3,double x4,double y1,double y2,double y3,double y4){
	Point ret;
	ret.x = ((x2-x1)*(y3*(x4-x3)+x3*(y3-y4))-(x4-x3)*(y1*(x2-x1)+x1*(y1-y2)))/((y2-y1)*(x4-x3)-(y4-y3)*(x2-x1));
	if(x1 != x2){
		ret.y = ((y2-y1)*ret.x+y1*(x2-x1)+x1*(y1-y2))/(x2-x1);
	}else{
		ret.y = ((y4-y3)*ret.x+y3*(x4-x3)+x3*(y3-y4))/(x4-x3);
	}
	return ret;
}

Point calc_Cross_Point(Point a,Point b,Point c,Point d){
	return calc_Cross_Point(a.x,b.x,c.x,d.x,a.y,b.y,c.y,d.y);
}

Polygon ConvexCut(Polygon g,Point a,Point b){

	Polygon ret;
	int N = g.size();

	for(int i = 0; i < g.size(); i++){
		Point A = g[i], B = g[(i+1)%N];
		if(ccw(a,b,A) != -1)ret.push_back(A);
		if(ccw(a,b,A)*ccw(a,b,B) == -1)ret.push_back(calc_Cross_Point(a,b,A,B));
	}

	return ret;
}

double calc_S(Polygon g){

	int N = g.size();
	double ret = 0;

	for(int i = 0; i < g.size(); i++){
		ret += cross(g[i],g[(i+1)%N]);
	}
	return ret/2.0;
}

int func(double x1,double y1,double x2, double y2, double xp, double yp){
	double naiseki,norm1,norm2,gaiseki;
	norm1 = sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
	norm2 = sqrt((xp-x1)*(xp-x1)+(yp-y1)*(yp-y1));
	naiseki = (xp-x1)*(x2-x1)+(yp-y1)*(y2-y1);
	gaiseki = (x2-x1)*(yp-y1)-(xp-x1)*(y2-y1);
	if(gaiseki > EPS){
		return 1;
	}else if(gaiseki < -EPS){
		return -1;
	}
	if(naiseki < -EPS){
		return 2;
	}

	if(norm1 < norm2){
		return -2;
	}
	return 0;
}


//★★直線ではなく、線分の交差判定★★
bool is_Cross(Line a,Line b){

	if(func(a.p[0].x,a.p[0].y,a.p[1].x,a.p[1].y,b.p[0].x,b.p[0].y)*
			func(a.p[0].x,a.p[0].y,a.p[1].x,a.p[1].y,b.p[1].x,b.p[1].y) <= 0 &&
			func(b.p[0].x,b.p[0].y,b.p[1].x,b.p[1].y,a.p[0].x,a.p[0].y)*
			func(b.p[0].x,b.p[0].y,b.p[1].x,b.p[1].y,a.p[1].x,a.p[1].y) <= 0){
		return true;
	}
	return false;
}

double calc_dist(Point A,Point B){

	return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}


int main(){

	scanf("%d",&N);

	Polygon L;

	double x,y;

	for(int i = 0; i < N; i++){

		scanf("%lf %lf",&x,&y);
		L.push_back(Point(x,y));
	}

	scanf("%d",&num_circle);


	for(int i = 0; i < num_circle; i++){
		scanf("%lf %lf",&C[i].x,&C[i].y);
	}

	double maximum = 0,tmp_dist;

	for(int i = 0; i < num_circle; i++){

		Polygon p = L;
		for(int k = 0; k < num_circle; k++){
			if(k == i)continue;

			Point from = (C[i]+C[k])/2; //中点
			Vector v1 = C[k]-C[i];
			Vector v2 = Vector(-v1.y,v1.x); //v1の法線ベクトル
			Point to = from+v2;
			p = ConvexCut(p,from,to);
			if(p.size() <= 2){
				break;
			}
		}
		if(p.size() <= 2)continue;

		tmp_dist = 0;

		for(int k = 0; k < p.size(); k++){

			tmp_dist = max(tmp_dist,calc_dist(C[i],p[k]));
		}
		maximum = max(maximum,tmp_dist);
	}

	printf("%.10lf\n",maximum);

	return 0;
}