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>
 
using namespace std;
 
#define MAX_N 100000
const double INF = 1e50;
 
typedef pair<double,double> P;
 
int N;
P A[MAX_N]; 
 
bool compare_y(const P &a,const 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() - 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(){
  sort(A,A+N);
  printf("%.8f\n",closest_pair(A,N));
}
 
int main(){
  cin >> N;
  for(int i = 0 ; i < N ; i++){
    scanf("%lf%lf",&A[i].first,&A[i].second);
  }
  solve();
  return 0;
}