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>
void swap(int* a, int* b) {
  int temp = *a;
  *a = *b;
  *b = temp;
}
void bsort(int n, int* ar) {
  int c = 1;
  while (c != 0) {
    c = 0;
    for (int i = 0; i < n - 1; i++) {
      if (ar[i] > ar[i + 1]) {
        swap(&ar[i], &ar[i + 1]);
        c++;
      }
    }
  }
}
int main(void) {
  int n;
  scanf("%d", &n);
  int ar[2 * n];
  for (int i = 0; i < 2 * n; i++) {
    scanf("%d", &ar[i]);
  }
  int x = 1;
  for (int i = 0; i < 2 * n; i++) {
    if (ar[0] != ar[i]) {
      x = 0;
      break;
    }
  }
  if (x == 1)
    printf("-1");
  else {
    bsort(2 * n, ar);
    for (int i = 0; i < 2 * n; i++) {
      printf("%d ", ar[i]);
    }
  }
  return 0;
}
