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 losn = 1e6 + 5;
const long long maxn = 1e5 + 5;
const long long minn = 1e3 + 5;
const long long tiny = 1e2 + 5;
const long long inf = 1e9;
const long long mod = 1e9 + 7;
const long long hmod = 4698571;
long long n;
long long arr[maxn];
int main() {
  cin >> n;
  for (long long i = 1; i <= n * 2; i++) {
    cin >> arr[i];
  }
  sort(arr + 1, arr + 1 + n * 2);
  long long l = 0;
  for (long long i = 1; i <= n; i++) {
    l += arr[i];
  }
  long long r = 0;
  for (long long i = n + 1; i <= n * 2; i++) {
    r += arr[i];
  }
  if (l != r) {
    for (long long i = 1; i <= n * 2; i++) {
      cout << arr[i] << " ";
    }
  } else
    cout << -1 << endl;
  return 0;
}
