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;
void change(int mas[], int n) {
  for (int i = 0; i < n; i++)
    for (int j = n; j < 2 * n; j++) {
      if (mas[j] != mas[i]) {
        swap(mas[i], mas[j]);
        return;
      }
    }
}
int main() {
  int mas[10000] = {0};
  int n, res;
  long long left = 0, right = 0;
  bool same = true;
  cin >> n;
  for (int i = 0; i < 2 * n; i++) {
    cin >> mas[i];
    if (i != 0 && mas[i] != mas[i - 1]) same = false;
    if (i < n)
      left += mas[i];
    else
      right += mas[i];
  }
  if (same) {
    cout << -1;
    return 0;
  } else if (left != right) {
    for (int i = 0; i < 2 * n; i++) cout << mas[i] << " ";
    return 0;
  } else {
    change(mas, n);
    for (int i = 0; i < 2 * n; i++) cout << mas[i] << " ";
    return 0;
  }
  return 0;
}
