#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#define p(s) cout<<(s)<<endl
#define REP(i,n,N) for(int i=n;i<N;i++)
#define RREP(i,n,N) for(int i=N-1;i>=n;i--)
#define CK(n,a,b) ((a)<=(n)&&(n)<(b))
#define F first
#define S second
typedef long long ll;
using namespace std;
const ll inf = 1e10;
int N;
ll x[1010],y[1010];
ll d[1010][1010];
bool visited[1010];

//答えが整数の時
bool C(ll mid){//ここに条件を書く

    queue<int> q;//ノード
    REP(i,0,N) visited[i]=false;    //初期位置を設定
    q.push(0);
    while(!q.empty()){
        int curNode = q.front();
        q.pop();
        visited[curNode] = true;
        REP(i,0,N){
            if(!visited[i]&&d[curNode][i]<=mid*mid) q.push(i);
        }
    }
    //lbが更新される不等式を書く
    return !visited[N-1];
}


ll nibutan(){

    ll lb=0,ub=inf;
    while(ub-lb>10){
        ll mid=(lb+ub)/2;
        if(mid%10!=0) mid=(mid/10+1)*10;
        if(C(mid)) lb=mid;
        else ub=mid;
        //cout<<lb<<" "<<ub<<endl;
    }
    return ub;
}

int main() {
    cin>>N;
    REP(i,0,N){
        cin>>x[i]>>y[i];
    }
    REP(i,0,N){
        REP(j,0,N){
            d[i][j]=(x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]);
        }
    }
    ll ans=nibutan();

    p(ans);

    return 0;
}