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 FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

typedef pair<double,double> pdd;
double INF=1000100010;
bool cmp_y(pdd a,pdd b){
    return a.second<b.second;
}
double closest_pair(pdd *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,cmp_y);
    vector<pdd> b;
    rep(i,n){
        if(abs(a[i].first-x)>=d) continue;
        rep(j,b.size()){
            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.pb(a[i]);
    }
    return d;
}
pdd a[100001];
int main(){
    int n;
    cin>>n;
    double xi,yi;
    rep(i,n){
        cin>>xi>>yi;
        a[i]={xi,yi};
    }
    sort(a,a+n);
    cout<<fixed<<setprecision(12)<<closest_pair(a,n)<<endl;
    return 0;
}
