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 long long int N = 1e7, MAXN = 5e3, delta = 46639, mod = 1e9 + 7,
                    oo = 1e16;
const long double PI = 3.141592653589793;
long long int a[N];
bool mark[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int n, dis = 0;
  cin >> n;
  for (long long int i = 0; i < 2 * n; i++) {
    cin >> a[i];
    if (!mark[a[i]]) dis++;
    mark[a[i]] = true;
  }
  if (dis == 1) return cout << (-1), 0;
  sort(a, a + (2 * n));
  for (long long int i = 0; i < 2 * n; i++) cout << a[i] << " ";
  return 0;
}
