Problem:
Score : 600 points

Problem Statement

Given are N points (x_i, y_i) in a two-dimensional plane.

Find the minimum radius of a circle such that all the points are inside or on it.

Constraints

2 \leq N \leq 50

0 \leq x_i \leq 1000

0 \leq y_i \leq 1000

The given N points are all different.

The values in input are all integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the minimum radius of a circle such that all the N points are inside or on it.

Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

2
0 0
1 0

Sample Output 1

0.500000000000000000

Both points are contained in the circle centered at (0.5,0) with a radius of 0.5.

Sample Input 2

3
0 0
0 1
1 0

Sample Output 2

0.707106781186497524

Sample Input 3

10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9

Sample Output 3

6.726812023536805158

If the absolute or relative error from our answer is at most 10^{-6}, the output will be considered correct.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <float.h>
#include <math.h>
#define N 50
float distance(float x1,float x2,float y1,float y2);
int main() {
	int n,i,j,k,breakflag;
	float r,r2,rMin;
	float x[N];
	float y[N];
	float mid_x;
	float mid_y;
	scanf("%d",&n);
	for(i=0;i<n;i++) {
		scanf("%f %f",&x[i],&y[i]);
	}
	rMin=FLT_MAX;
	breakflag=0;
	for(i=0;i<n;i++) {
		for(j=0;j<n;j++) {
			breakflag=0;
			mid_x=(x[i]+x[j])/2;
			mid_y=(y[i]+y[j])/2;
			r=distance(x[j],mid_x,y[j],mid_y);
			for(k=0;k<n;k++) {
				r2=distance(x[k],mid_x,y[k],mid_y);
				if(r2>r) {
					breakflag=1;
					break;
				}
			}
			if(breakflag==1) {
				breakflag=0;
				continue;
			} else {
				if(rMin>r) {
					rMin=r;
				}
			}
		}
	}
	printf("%f",rMin);
	return 0;
}

float distance(float x1,float x2,float y1,float y2){
	float res;
	res=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
	return res;
}