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 sort(int left, int right);
int a[2002];
int main(void) {
  int n, i, m, x, y = 0;
  scanf("%d", &n);
  for (i = 0; i < 2 * n; i++) {
    scanf("%d", &a[i]);
  }
  for (x = a[0], i = 0; i < 2 * n; i++) {
    if (x != a[i]) y = 1;
  }
  if (y) {
    sort(0, 2 * n - 1);
    for (i = 0; i < 2 * n; i++) {
      printf("%d ", a[i]);
    }
  } else
    printf("-1");
  return 0;
}
void sort(int left, int right) {
  if (left >= right) {
    return;
  }
  int i = left, j = right, key = a[left];
  while (i < j) {
    while (i < j && key <= a[j]) {
      j--;
    }
    a[i] = a[j];
    while (i < j && key >= a[i]) {
      i++;
    }
    a[j] = a[i];
  }
  a[i] = key;
  sort(left, i - 1);
  sort(i + 1, right);
}
