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>
int main() {
  int t, n, i, j, sum1, sum2, flag = 0, k = 0;
  int a[5050];
  scanf("%d", &n);
  for (i = 1; i <= 2 * n; i++) {
    scanf("%d", &a[i]);
    if (i <= n)
      sum1 += a[i];
    else
      sum2 += a[i];
  }
  for (i = 1; i <= 2 * n - 1; i++) {
    if (a[i] == a[i + 1]) {
      flag++;
    }
  }
  if (flag == n * 2 - 1)
    printf("-1\n");
  else {
    for (i = 1; i <= n; i++) {
      for (j = n + 1; j <= 2 * n; j++) {
        if (a[i] != a[j] && (sum1 - a[i] + a[j]) != (sum2 - a[j] + a[i])) {
          t = a[i];
          a[i] = a[j];
          a[j] = t;
          k = 1;
          break;
        }
      }
      if (k == 1) break;
    }
    for (i = 1; i <= 2 * n; i++) printf("%d ", a[i]);
  }
  return 0;
}
