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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t, n, i, f = 0, s2 = 0, s1 = 0;
  t = 1;
  while (t--) {
    cin >> n;
    long long int ar[2 * n];
    for (i = 0; i < 2 * n; i++) cin >> ar[i];
    for (i = 0; i < 2 * n - 1; i++) {
      if (ar[i] != ar[i + 1]) {
        f = 1;
        break;
      }
    }
    for (i = 0; i < n; i++) {
      s1 += ar[i];
      s2 += ar[n + i];
    }
    if (f == 0)
      cout << "-1\n";
    else {
      sort(ar, ar + 2 * n);
      for (i = 0; i < 2 * n; i++) cout << ar[i] << " ";
      cout << "\n";
    }
    s1 = s2 = f = 0;
  }
}
