problem

Given $ N $ different natural numbers $ a_i $. I decided to make a pair by choosing a different natural number from the given natural numbers. Output one pair that can be created with a value difference that is a multiple of $ N -1 $.

It should be noted that such a pair always exists.





Example

Input

5
1 2 4 7 10


Output

2 10
#include <iostream>
#include <vector>
#include <map>
#include <utility>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <set>
#include <unordered_map>
#include <unordered_set>

#define REP(i,a,n) for(int i=a;i<n;++i)
#define rep(i,n) REP(i,0,n)
#define REV(i,a,n) for(int i=n;i>=a;--i)
#define all(e) e.begin(),e.end()
#define rall(e) e.rbegin(),e.rend()
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define mod 1000000007
#define show(n) cerr<<#n<<" = "<<n<<endl
#define shows(n) for(auto z:n){cerr<<z<<", ";}cerr<<endl

#define yes cout << "YES" << endl
#define no cout << "NO" << endl

using namespace std;

using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using pint = pair<int,int>;

const int INF = (1 << 30);

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vi vec(n);
    rep(i, n){
        cin >> vec[i];
    }

    rep(i, n){
        rep(j, n){
            //show(abs(vec[i] - vec[j]));
            if(abs(vec[i] - vec[j]) != 0){
                if(abs(vec[i] - vec[j]) % (n - 1) == 0){
                    cout << vec[i] << ' ' << vec[j] << endl;
                    return 0;
                }
            }
            
        }
    }

    return 0;
}
