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 <bits/stdc++.h>
 
using namespace std;
 
//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
 
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
 
//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second
 
//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);

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

  int N;
  cin >> N;
  VI xs(N);
  REP(i,N) cin >> xs[i];

  REP(i,N) REP(j,i){
	if((xs[i] - xs[j]) % (N-1) == 0){
	  cout << xs[i] << " " << xs[j] << endl;
	  return 0;
	}
  }
  
  return 0;
}

