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 <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <functional>
#include <numeric>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <fstream>
#include <bitset>
#include <time.h>
#include <tuple>
#include <iomanip>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef vector<ll> V;
typedef complex<double> Point;

#define PI acos(-1.0)
#define EPS 1e-10
const ll INF = 1e12;
const ll MOD = 1e9 + 7;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define rep(i,N) for(int i=0;i<(N);i++)
#define ALL(s) (s).begin(),(s).end()
#define EQ(a,b) (abs((a)-(b))<EPS)
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )
#define fi first
#define se second
#define N_SIZE (1LL << 20)
#define NIL -1

ll sq(ll num) { return num*num; }
ll mod_pow(ll x, ll n) {
	if (n == 0)return 1;
	if (n == 1)return x%MOD;
	ll res = sq(mod_pow(x, n / 2));
	res %= MOD;
	if (n % 2 == 1) {
		res *= x;
		res %= MOD;
	}
	return res;
}
ll mod_add(ll a, ll b) { return (a + b) % MOD; }
ll mod_sub(ll a, ll b) { return (a - b + MOD) % MOD; }
ll mod_mul(ll a, ll b) { return a*b % MOD; }

ll n;
ll a[1010];

int main() {
	cin >> n;
	rep(i, n)cin >> a[i];
	sort(a, a + n);
	rep(i, n) {
		FOR(j, i + 1, n) {
			if ((a[j] - a[i]) % (n - 1) == 0) {
				cout << a[j] << " " << a[i] << endl;
				return 0;
			}
		}
	}
}
