Problem: 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

Solution (in PYTHON3):

Solution:
# -*- coding: utf-8 -*-
# F: 2次元での三分探査
# 最小包含円を探索する（中心と半径）

import math

N = int(input())
X, Y = [], []
for _ in range(N):
    x, y = map(int, input().split())
    X.append(x)
    Y.append(y)


def calc_r(x0, y0):
    # 全ての点の中で最も遠いところまでの距離を返す
    res = 0
    for i in range(N):
        res = max(res, ((X[i] - x0)**2 + (Y[i] - y0)**2))

    return res


# lb_x, ub_x = min(X), max(X)
lb_x, ub_x = 0.0, 1000.0


def g(x):
    # lb_y, ub_y = min(Y), max(Y)
    lb_y, ub_y = 0.0, 1000.0

    # yを探索する
    for _ in range(100):
        y1 = lb_y + (ub_y - lb_y) / 3
        y2 = ub_y - (ub_y - lb_y) / 3

        if calc_r(x, y1) > calc_r(x, y2):
            lb_y = y1
        else:
            ub_y = y2
    return calc_r(x, lb_y)


for _ in range(100):
    # あるxにおいて半径が最小となる円を求める
    x1 = lb_x + (ub_x - lb_x) / 3
    x2 = ub_x - (ub_x - lb_x) / 3

    if g(x1) > g(x2):
        lb_x = x1
    else:
        ub_x = x2

print(math.sqrt(g(lb_x)))