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}.

Examples

Input

2
0 0
1 0


Output

0.500000000000000000


Input

3
0 0
0 1
1 0


Output

0.707106781186497524


Input

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


Output

6.726812023536805158
# -*- coding: utf-8 -*-
import sys
import cmath
sys.setrecursionlimit(10**9)
INF=10**18
MOD=10**9+7
eps=1e-8
input=lambda: sys.stdin.readline().rstrip()
YesNo=lambda b: bool([print('Yes')] if b else print('No'))
YESNO=lambda b: bool([print('YES')] if b else print('NO'))
int1=lambda x:int(x)-1

def main():
    N=int(input())
    x,y=[0]*N,[0]*N
    for i in range(N):
        x[i],y[i]=map(int,input().split())
    
    def get_Intersection(c1,c2,R):
        mid=(c1+c2)/2
        if R<abs(c1-mid):
            return None,None
        d=(R**2-abs(c1-mid)**2)**0.5
        p=cmath.polar((c1-c2)*1j)
        v=cmath.rect(d,p[1])
        i1=mid+v
        i2=mid-v
        return i1,i2
    
    def nibutan(ok,ng):
        for _ in range(100):
            mid = (ok + ng) / 2
            if solve(mid):
                ok = mid
            else:
                ng = mid
        return ok
    
    def solve(mid):
        for i in range(N):
            for j in range(N):
                if i==j:
                    continue
                c1=x[i]+y[i]*1j
                c2=x[j]+y[j]*1j
                i1,i2=get_Intersection(c1,c2,mid)
                if i1 is None:
                    continue
                for k in range(N):
                    i3=x[k]+y[k]*1j
                    if abs(i3-i1)>mid+eps:
                        break
                else:
                    return True
                for k in range(N):
                    i3=x[k]+y[k]*1j
                    if abs(i3-i2)>mid+eps:
                        break
                else:
                    return True
        return False
    
    print(nibutan(2000,0))
    


    

if __name__ == '__main__':
    main()
