Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n; double bx, by; pair<double, double> p[100009];
int main() {
	scanf("%d", &n);
	double sq = sin(1.234), cq = cos(1.234);
	for(int i = 0; i < n; i++) {
		scanf("%lf%lf", &bx, &by);
		double ex = bx * cq - by * sq;
		double ey = bx * sq + by * cq;
		p[i] = make_pair(ex, ey);
	}
	sort(p, p + n);
	double ret = 1.0e+10;
	for(int i = 0; i < n; i++) {
		for(int j = i + 1; j < n; j++) {
			double dx = p[i].first - p[j].first;
			if(dx * dx > ret) break;
			double dy = p[i].second - p[j].second;
			double dist = dx * dx + dy * dy;
			if(ret > dist) ret = dist;
		}
	}
	printf("%.15lf\n", sqrt(ret));
	return 0;
}