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;
const long long N = 2e5 + 5, MOD = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long a[N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  n *= 2;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  long long sum = 0, tot = 0;
  for (long long i = n / 2 + 1; i <= n; i++) {
    sum += a[i];
  }
  for (long long i = 1; i <= n / 2; i++) tot += a[i];
  if (tot != sum) {
    for (long long i = 1; i <= n; i++) cout << a[i] << " ";
  } else
    cout << -1;
  return 0;
}
