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 a[1000], n, b[1000], x = 0, sum, y, z;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    y = y + a[i];
  }
  for (int j = 0; j < n; j++) {
    cin >> b[j];
    z = z + b[j];
  }
  if (z == y) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (a[i] != b[j] && x < 1) {
          sum = b[j];
          b[j] = a[i];
          a[i] = sum;
          x++;
        }
      }
    }
  }
  if (x == 0 && z == y)
    cout << -1;
  else if (x > 0 && z == y) {
    for (int i = 0; i < n; i++) cout << a[i] << ' ';
    for (int j = 0; j < n; j++) cout << b[j] << ' ';
  } else {
    for (int i = 0; i < n; i++) cout << a[i] << ' ';
    for (int j = 0; j < n; j++) cout << b[j] << ' ';
  }
}
