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;
}
//??????????????¨????????§????±?????°???????????????????????????????????????????
//?±????????????????????????????x???????????§d????????¢????°???????????????¨???
//????????????????????§???????±?1000000??§??????????????????????????????
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 < (int)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;
}