Problem: 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

Solution (in CPP):

Solution:
#include <cassert>// c
#include <iostream>// io
#include <iomanip>
#include <fstream>
#include <sstream>
#include <vector>// container
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <stack>
#include <algorithm>// other
#include <complex>
#include <numeric>
#include <functional>
using namespace std;

typedef int32_t i32;typedef int64_t i64;typedef i64 ll;typedef uint32_t uint;typedef uint64_t ull;

#define ALL(c) (begin(c)),(end(c))
#define REP(i,n) FOR(i,0,n)
#define REPr(i,n) FORr(i,0,n)
#define FOR(i,l,r) for(int i=(int)(l);i<(int)(r);++i)
#define FORr(i,l,r) for(int i=(int)(r)-1;i>=(int)(l);--i)
#define EACH(it,o) for(__typeof((o).begin()) it = (o).begin(); it != (o).end(); ++it)
#define IN(l,v,r) ((l)<=(v) && (v)<(r))

//debug
#define DUMP(x)  cerr << #x << " = " << (x)
#define DUMPLN(x)  DUMP(x) <<endl
#define DEBUG(x) DUMP(x) << LINE() << " " << __FILE__
#define DEBUGLN(x) DEBUG(x)<<endl
#define LINE()    cerr<< " (L" << __LINE__ << ")"
#define LINELN()    LINE()<<endl
#define CHECK(exp,act)  if(exp!=act){DUMPLN(exp);DEBUGLN(act);}
#define STOP(e)  CHECK(e,true);if(!(e)) exit(1);

class range {
private:
    struct Iter{
        int v;
        int operator*(){return v;}
        bool operator!=(Iter& itr) {return v < itr.v;}
        void operator++() {++v;}
    };
    Iter i, n;
public:
    range(int n) : i({0}), n({n}) {}
    range(int i, int n) : i({i}), n({n}) {}
    Iter& begin() {return i;}
    Iter& end() {return n;}
};

//output
template<typename T> ostream& operator << (ostream& os, const vector<T>& as){REP(i,as.size()){if(i!=0)os<<" "; os<<as[i];}return os;}
template<typename T> ostream& operator << (ostream& os, const vector<vector<T> >& as){REP(i,as.size()){if(i!=0)os<<endl; os<<as[i];}return os;}
template<typename T> ostream& operator << (ostream& os, const set<T>& ss){EACH(a,ss){if(a!=ss.begin())os<<" "; os<<a;}return os;}
template<typename T1,typename T2> ostream& operator << (ostream& os, const pair<T1,T2>& p){os<<p.first<<" "<<p.second;return os;}
template<typename K,typename V> ostream& operator << (ostream& os, const map<K,V>& m){bool isF=true;EACH(p,m){if(!isF)os<<endl;os<<p;isF=false;}return os;}

//input
char tmp[1000];
#define nextInt(n) scanf("%d",&n)
#define nextLong(n) scanf("%lld",&n) //I64d
#define nextDouble(n) scanf("%lf",&n) 
#define nextChar(n) scanf("%c",&n)
#define nextString(n) scanf("%s",tmp);n=tmp

// values
template<class T> T INF(){assert(false);};
template<> int INF<int>(){return 1<<28;};
template<> ll INF<ll>(){return 1LL<<58;};
template<> double INF<double>(){return 1e16;};
template<> long double INF<long double>(){return 1e16;};

template<typename T> T pmod(T v,T M){return (v%M+M)%M;}

namespace _double_tmpl{
    typedef long double D;const D EPS = 1e-8;
    
    static constexpr D Ae=0;
    D A(D a,D b){return a+b;}D Ainv(D a){return -a;}
    D S(D a,D b){return A(a,Ainv(b));}
    
    static constexpr D Me=1;
    D M(D a,D b){return a*b;}D Minv(D a){return 1.0/a;};

    int sig(D a,D b=0){return a<b-EPS?-1:a>b+EPS?1:0;}
    bool eq(D a,D b){ return sig(abs(a-b))==0;}

}
using namespace _double_tmpl;


namespace Ps{
    // using namespace _double_tmpl;
    typedef complex<D> P,Vec;

#define X real()
#define Y imag()
    istream& operator >> (istream& is,complex<D>& p){
        D x,y;is >> x >> y;p=P(x,y);return is;
    }

    bool eq(P a,P b){return eq(a.X,b.X) and eq(a.Y,b.Y);}
    // a×b
    D cross(const Vec& a,const Vec& b){return imag(conj(a)*b);}
    // a・b
    D dot(const Vec&a,const Vec& b) {return real(conj(a)*b);}

    int ccw(const P& a,P b,P c){
        b -= a; c -= a;
        if (cross(b,c) > EPS)   return +1;       // counter clockwise
        if (cross(b,c) < -EPS)   return -1;       // clockwise
        if (dot(b,c) < 0)     return +2;       // c--a--b on line
        if (norm(b) < norm(c)) return -2;       // a--b--c on line
        return 0;
    }

    bool compY (const P& a,const P& b);
    D closestPair(vector<P>& ps,int l,int r){
        if(r-l<2)return INF<D>();
        D res=min(closestPair(ps,l,(l+r)/2),closestPair(ps,(l+r)/2,r));
        
        vector<P> ips;
        FOR(i,l,r)if(abs(ps[i].X - ps[(l+r)/2].X)<res)ips.push_back(ps[i]);
        sort(ALL(ips),compY);        
        
        REP(i,ips.size())for(int j=i-10;j<i;j++)if(j>=0)
            res=min(res,abs(ips[i]-ips[j]));

        return res;
    }
    bool compY (const P& a,const P& b){return a.Y != b.Y ? a.Y < b.Y : a.X < b.X;}
}
using namespace Ps;

namespace std{

    bool compX (const P& a,const P& b){return a.X != b.X ? a.X < b.X : a.Y < b.Y;}
    bool compY (const P& a,const P& b){return a.Y != b.Y ? a.Y < b.Y : a.X < b.X;}
    bool operator < (const P& a,const P& b){return compX(a,b);}
    bool operator == (const P& a,const P& b){return eq(a,b);}
};

stringstream ss;
class Main{
    public:
    void run(){
        int n;cin >> n;
        vector<P> ps;
        REP(i,n) {
            P p;cin >> p;
            ps.push_back(p);
        }
        sort(ALL(ps));
        cout << closestPair(ps,0,n)<<endl;
    }
};

int main(){
    cout <<fixed<<setprecision(20);
    cin.tie(0);
    ios::sync_with_stdio(false);
    Main().run();
    return 0;
}