Problem: Find the diameter of a convex polygon g. In other words, find a pair of points that have maximum distance between them.

Constraints

* 3 ≤ n ≤ 80000
* -100 ≤ xi, yi ≤ 100
* No point in the g will occur more than once.

Input


n
x1 y1
x2 y2
:
xn yn


The first integer n is the number of points in g.

In the following lines, the coordinate of the i-th point pi is given by two real numbers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Each value is a real number with at most 6 digits after the decimal point.

Output

Print the diameter of g in a line. The output values should be in a decimal fraction with an error less than 0.000001.

Examples

Input

3
0.0 0.0
4.0 0.0
2.0 2.0


Output

4.00


Input

4
0.0 0.0
1.0 0.0
1.0 1.0
0.0 1.0


Output

1.414213562373

Solution (in PYTHON):

Solution:
from math import sqrt
n = input()
ps = [map(float, raw_input().split()) for i in xrange(n)]
ps.sort()

def cross(a, b, c, d):
    return (b[0]-a[0])*(d[1]-c[1]) - (b[1]-a[1])*(d[0]-c[0])
def convex_hull(ps):
    qs = []
    n = len(ps)
    for p in ps:
        while len(qs)>1 and cross(qs[-1], qs[-2], qs[-1], p) >= 0:
            qs.pop()
        qs.append(p)
    t = len(qs)
    for i in xrange(n-2, -1, -1):
        p = ps[i]
        while len(qs)>t and cross(qs[-1], qs[-2], qs[-1], p) >= 0:
            qs.pop()
        qs.append(p)
    return qs

def dist(a, b):
    return sqrt((a[0]-b[0])**2 + (a[1]-b[1])**2)
def rotating_calipers(ps):
    qs = convex_hull(ps)
    n = len(qs)
    if n == 2:
        return dist(qs[0], qs[1])
    i = j = 0
    for k in xrange(n):
        if qs[k] < qs[i]: i = k
        if qs[j] < qs[k]: j = k
    res = 0
    si = i; sj = j
    while i != sj or j != si:
        res = max(res, dist(qs[i], qs[j]))
        if cross(qs[i], qs[i-n+1], qs[j], qs[j-n+1]) < 0:
            i = (i + 1) % n
        else:
            j = (j + 1) % n
    return res
print "%.09f" % rotating_calipers(ps)