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;
const int N = 1e6 + 5;
int a[N];
int main() {
  int n, m, i, j, k, x, y, z;
  long long s1 = 0, s2 = 0;
  scanf("%d", &n);
  for (i = 1; i <= 2 * n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + 2 * n + 1);
  for (i = 1; i <= n; i++) s1 += a[i];
  for (i = n + 1; i <= 2 * n; i++) s2 += a[i];
  if (s1 == s2) {
    printf("-1\n");
    return 0;
  }
  for (i = 1; i <= 2 * n; i++) printf("%d ", a[i]);
}
