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
#include <bits/stdc++.h>
#define MAX 100001
using namespace std;

typedef pair<double,double> P;
const int INF=1000000000;
int n;
P v[MAX];
 
bool compare_y(P a, P b) {
  return a.second < b.second;
}
 
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()-1-j].first;
      double dy=a[i].second-b[b.size()-1-j].second;
      if(dy>=d) break;
      d=min(d, sqrt(dx*dx+dy*dy));
    }
    b.push_back(a[i]);
  }
  return d;
}

int main() {
  cin >> n;
  for(int i=0; i<n; i++) {
    double x,y;
    cin >> x >> y;
    v[i]=P(x,y);
  }
  sort(v,v+n);
  double ans=closest_pair(v,n);
  printf("%.10f\n", ans);
  return 0;
}
