#include<bits/stdc++.h>
using namespace std;
ifstream fin("cmap.in");
ofstream fout("cmap.out");

const int DIM = 1e5 + 5;

int N;
pair<int,int> arr[DIM], aux[DIM];

inline long long dist( pair<int,int> A, pair<int,int> B ){
    return 1LL * (A.first - B.first) * (A.first - B.first) +
           1LL * (A.second - B.second) * (A.second - B.second);
}

long long solve( int st, int dr ){
    if( dr - st == 1 ){
        long long ans = dist( arr[st], arr[dr] );
        if( arr[st].second > arr[dr].second )
            swap( arr[st], arr[dr] );
        return ans;
    }
    if( dr - st == 2 ){
        long long ans = dist( arr[st], arr[st + 1] );
        ans = min( ans, dist( arr[st], arr[dr] ) );
        ans = min( ans, dist( arr[st + 1], arr[dr] ) );
        if( arr[st].second > arr[st + 1].second )
            swap( arr[st], arr[st + 1] );
        if( arr[st].second > arr[dr].second )
            swap( arr[st], arr[dr] );
        if( arr[st + 1].second > arr[dr].second )
            swap( arr[st + 1], arr[dr] );
        return ans;
    }
    int mid = ( st + dr ) >> 1;
    long long ans = solve( st, mid );
    ans = min( ans, solve( mid + 1, dr ) );

    int p1 = st, p2 = mid + 1, k = st - 1;
    while( p1 <= mid && p2 <= dr ){
        if( arr[p1].second < arr[p2].second )
            aux[++k] = arr[p1++];
        else
            aux[++k] = arr[p2++];
    }
    while( p1 <= mid )
        aux[++k] = arr[p1++];
    while( p2 <= dr )
        aux[++k] = arr[p2++];
    for( int i = st; i <= dr; i++ )
        arr[i] = aux[i];

    k = 0;
    for( int i = st; i <= dr; i++ )
        if( (long long)( abs( (arr[mid].first - arr[i].first) ) ) <= ans )
            aux[++k] = arr[i];

    for( int i = 1; i <= k; i++ )
        for( int j = i + 1; j <= i + 7 && j <= k; j++ )
            ans = min( ans, dist( aux[i], aux[j] ) );

    return ans;
}


int main(){
    fin >> N;
    for( int i = 1; i <= N; i++ )
        fin >>  arr[i].first >> arr[i].second;
    sort( arr + 1, arr + N + 1 );
    fout << setprecision(6) << fixed << sqrt( 1.0 * solve( 1, N ) ) << "\n";
    return 0;
}