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
#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

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

  int n;
  cin >> n;
  VI a(n);
  REP(i, n) cin >> a[i];

  REP(i, n) FOR(j, i+1, n) {
    if(abs(a[i]-a[j])%(n-1) == 0) {
      cout << a[i] << " " << a[j] << endl;
      return 0;
    }
  }

  return 0;
}

