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 <iostream>
#include <iomanip>
#include <algorithm>
#include <complex>
#include <utility>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <tuple>
#include <cmath>
#include <bitset>
#include <cctype>
#include <set>
#include <map>
#include <unordered_map>
#include <numeric>
#include <functional>
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(ll i=ll(a);i<ll(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define PRINT(V) cout << V << "\n"
#define SORT(V) sort((V).begin(),(V).end())
#define RSORT(V) sort((V).rbegin(), (V).rend())
using namespace std;
using ll = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline void Yes(bool condition){ if(condition) PRINT("Yes"); else PRINT("No"); }
template<class itr> void cins(itr first,itr last){
    for (auto i = first;i != last;i++){
        cin >> (*i);
    }
}
template<class itr> void array_output(itr start,itr goal){
    string ans = "",k = " ";
    for (auto i = start;i != goal;i++) ans += to_string(*i)+k;
    if (!ans.empty()) ans.pop_back();
    PRINT(ans);
}
ll gcd(ll a, ll b) {
    return a ? gcd(b%a,a) : b;
}
const ll INF = 1e15;
const ll MOD = 1000000007;
const ll MOD2 = 998244353;
typedef pair<ll,ll> P;
typedef pair<double,double> point;
const ll MAX = 2010;
constexpr ll nx[8] = {1,0,-1,0,-1,-1,1,1};
constexpr ll ny[8] = {0,1,0,-1,-1,1,-1,1};
ll n;
point g[100010];
bool cmp_y(point a,point b){
    return a.second < b.second;
}
double closest_pair(point *a,ll n){
    if (n <= 1) return INF;
    ll 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<point> b;
    rep(i,n){
        if (fabs(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;
            chmin(d,sqrt(dx*dx+dy*dy));
        }
        b.push_back(a[i]);
    }
    return d;
}
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    double x,y;
    cin >> n;
    rep(i,n){
        cin >> x >> y;
        g[i] = point(x,y);
    }
    sort(g,g+n);
    printf("%.10f\n",closest_pair(g,n));
}

