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
def f_enclose_all():
    # 参考: https://maspypy.com/atcoder-参加感想-2019-01-10abc-151
    from itertools import combinations
    N = int(input())
    Pos_complex = [complex(*map(int, input().split())) for j in range(N)]  # 複素数で持つ

    def find_circumcenter(z0, z1, z2):
        """複素数で表現される3点に対し，それらの外心を求める"""
        z1 -= z0
        z2 -= z0  # z0を基準に考える

        if abs((z1 * z2.conjugate()).imag) < 0.5:
            return None  # 3点が同一線上にあるので，外心を定義できない

        numerator = z1 * z2 * (z1 - z2).conjugate()
        denominator = z1.conjugate() * z2 - z1 * z2.conjugate()
        return (numerator / denominator) + z0

    # 2点選んで中点の座標を求めると，それが中心の候補
    candidate = [(z1 + z2) / 2 for z1, z2 in combinations(Pos_complex, 2)]
    # 3点選んで外心の座標を求めると，それが中心の候補
    candidate.extend([find_circumcenter(*comb) for comb in combinations(Pos_complex, 3)])

    # 中心の候補と各点への距離の最大値が包含円の半径．それを最小にする中心を探す
    ans = min((max(abs(z - c) for z in Pos_complex) for c in candidate if c is not None))
    return round(ans, 10)

print(f_enclose_all())