#include <fstream>
#include <algorithm>
#include <cmath>
#include <iomanip>
#define x first
#define y second

using namespace std;

ifstream fin("cmap.in");
ofstream fout("cmap.out");

int n, i, k;
pair<int, int> v[100001], a[100001], Y[100001];

int dist1(int i, int j){
    return (v[i].x-v[j].x)*(v[i].x-v[j].x) + (v[i].y-v[j].y)*(v[i].y-v[j].y);
}

int dist2(int i, int j){
    return (a[i].x-a[j].x)*(a[i].x-a[j].x) + (a[i].y-a[j].y)*(a[i].y-a[j].y);
}

int sol(int st, int dr){
    if(dr == st+1){
        return dist1(st, dr);
    }
    if(dr == st+2){
        return min(dist1(st, st+1), min(dist1(st, st+2), dist1(st+1, st+2)));
    }
    int mid = (st+dr)/2;
    int d1 = sol(st, mid);      /// solutia din stanga
    int d2 = sol(mid+1, dr);    /// solutia din dreapta
    int d = min(d1, d2);        /// cea mai buna solutie care se afla ba in stanga, ba in dreapta
    /// dupa ce mi-am facut treaba in fiecare jumatate
    /// fiecare jumatate este sortata dupa y
    /// deci le pot interclasa pentru a sorta intregul sir st...dr dupa y
    int j = mid+1, t = 0, i=st;
    while(i<=mid && j<=dr){
        if(v[i].y < v[j].y)
            Y[++t] = v[i++];
        else
            Y[++t] = v[j++];
    }
    for(;i<=mid;i++)
        Y[++t] = v[i];
    for(;j<=dr;j++)
        Y[++t] = v[j];
    int grup = 0;
    for(i=st;i<=dr;i++){
        v[i] = Y[i-st+1];
        if(dist1(i, mid) <= d)   /// imi aleg ca referinta punctul mid
            a[++grup] = v[i];
    }
    for(i=1;i<=grup-7;i++)
        for(j=i+1;j<=i+7;j++)
            d = min(d, dist2(i, j));
    return d;
}

int main(){
    fin>>n;
    for(i=1;i<=n;i++)
        fin>>v[i].x>>v[i].y;
    sort(v+1, v+n+1);
    fout<<setprecision(6)<<fixed<<(double)sqrt(sol(1, n));
    return 0;
}