You're given an array a of length 2n. Is it possible to reorder it in such way so that the sum of the first n elements isn't equal to the sum of the last n elements?

Input

The first line contains an integer n (1 ≤ n ≤ 1000), where 2n is the number of elements in the array a.

The second line contains 2n space-separated integers a_1, a_2, …, a_{2n} (1 ≤ a_i ≤ 10^6) — the elements of the array a.

Output

If there's no solution, print "-1" (without quotes). Otherwise, print a single line containing 2n space-separated integers. They must form a reordering of a. You are allowed to not change the order.

Examples

Input


3
1 2 2 1 3 1


Output


2 1 3 1 1 2

Input


1
1 1


Output


-1

Note

In the first example, the first n elements have sum 2+1+3=6 while the last n elements have sum 1+1+2=4. The sums aren't equal.

In the second example, there's no solution.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ldb = long double;
using pa = pair<ll, ll>;
using vec = vector<ll>;
const int MOD = 1e9 + 7, INF = INT_MAX, N = 1e5 + 10;
const double PI = acos(-1);
const ll LINF = LLONG_MAX;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll a;
  cin >> a;
  vector<ll> q(2 * a);
  for (ll i = 0; i < 2 * a; i++) {
    cin >> q[i];
  }
  sort(q.begin(), q.end());
  ll s = 0;
  for (ll i = 0; i < a; i++) s += q[i];
  for (ll i = a; i < 2 * a; i++) s -= q[i];
  if (s == 0)
    cout << -1 << "\n";
  else {
    for (auto el : q) {
      cout << el << " ";
    }
  }
  return 0;
}
