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;

//#################################################################
//最近点対 O(n log n)

namespace closest_pair{

typedef pair<double, double> P;

const int MAX_N = 1000000;
const double INF = 1e77;

int N;
P A[MAX_N];

bool compare_y(P a, P b){
	return a.second < b.second;
}

//配列aはsortしなければいけない！！
double closest_pair(P *a, int n){
	if(n <= 1) return INF;
	int m = n / 2;
	double x = a[m].first;
	double d = min(closest_pair(a, m), closest_pair(a + m, n - m));
	inplace_merge(a, a + m, a + n, compare_y);

	vector<P> b;
	for(int i=0;i<n;i++){
		if(fabs(a[i].first - x) >= d) continue;
		for(int j=0;j<b.size();j++){
			double dx = a[i].first - b[b.size() - j - 1].first;
			double dy = a[i].second - b[b.size() - j - 1].second;
			if(dy >= d) break;
			d = min(d, sqrt(dx * dx + dy * dy));
		}
		b.push_back(a[i]);
	}
	return d;
}

  void solve() {
    cin >> N;
    for(int i = 0; i < N; ++i) {
      cin >> A[i].first >> A[i].second;
    }
    sort(A, A+N);
    printf("%.10f\n", closest_pair(A, N));
  }

}

int main() {
  closest_pair::solve();
  return 0;
}