For given n points in metric space, find the distance of the closest points.

Constraints

* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100

Input


n
x0 y0
x1 y1
:
xn-1 yn-1


The first integer n is the number of points.

In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.

Output

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

Examples

Input

2
0.0 0.0
1.0 0.0


Output

1.000000


Input

3
0.0 0.0
2.0 0.0
1.0 1.0


Output

1.41421356237
from math import sqrt
n = input()
ps = [map(float, raw_input().split()) for i in xrange(n)]
ps.sort()
INF = 10**9

def cp_rec(ps, i, n):
    if n <= 1:
        return INF, [ps[i]]
    m = n/2
    x = ps[i+m][0]
    d1, qs1 = cp_rec(ps, i, m)
    d2, qs2 = cp_rec(ps, i+m, n-m)
    d = min(d1, d2)
    # merge
    qs = [None]*n
    s = t = idx = 0
    while s < m and t < n-m:
        if qs1[s][1] < qs2[t][1]:
            qs[idx] = qs1[s]; s += 1
        else:
            qs[idx] = qs2[t]; t += 1
        idx += 1
    while s < m:
        qs[idx] = qs1[s]; s += 1
        idx += 1
    while t < n-m:
        qs[idx] = qs2[t]; t += 1
        idx += 1
    b = []
    for i in xrange(n):
        ax, ay = q = qs[i]
        if abs(ax - x) >= d:
            continue
        for j in xrange(len(b)-1, -1, -1):
            dx = ax - b[j][0]
            dy = ay - b[j][1]
            if dy >= d: break
            d = min(d, sqrt(dx**2 + dy**2))
        b.append(q)
    return d, qs
def closest_pair(ps):
    n = len(ps)
    return cp_rec(ps, 0, n)[0]
print "%.09f" % closest_pair(ps)