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 int MAXN = 200005;
const int MOD = 1000000007;
int N;
int A[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = (0); i <= (2 * N - 1); i++) cin >> A[i];
  sort(A, A + 2 * N);
  long long s = 0;
  for (int i = (0); i <= (N - 1); i++) s += A[i];
  for (int i = (N); i <= (2 * N - 1); i++) s -= A[i];
  if (s == 0)
    cout << "-1\n";
  else {
    for (int i = (0); i <= (2 * N - 1); i++)
      cout << A[i] << " \n"[i == 2 * N - 1];
  }
}
