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() {
  int t;
  cin >> t;
  int n = 2 * t;
  long long arr[n], i, j;
  for (i = 0; i < n; i++) cin >> arr[i];
  long long temp = arr[0];
  int flag = 1;
  for (i = 1; i < n; i++) {
    if (arr[i] != temp) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    cout << "-1";
    return 0;
  }
  long long sum_1 = 0, sum_2 = 0;
  for (i = 0; i < t; i++) {
    sum_1 += arr[i];
  }
  for (i = t; i < n; i++) {
    sum_2 += arr[i];
  }
  if (sum_1 != sum_2) {
    for (i = 0; i < n - 1; i++) cout << arr[i] << " ";
    cout << arr[i];
    return 0;
  }
  flag = 0;
  for (j = 0; j < t; j++) {
    temp = arr[j];
    for (i = t; i < n; i++) {
      if (arr[i] != temp) {
        arr[j] = arr[i];
        arr[i] = temp;
        flag = 1;
        break;
      }
    }
    if (flag) break;
  }
  for (i = 0; i < n - 1; i++) cout << arr[i] << " ";
  cout << arr[i];
}
