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 M = 1000003;
struct Compare {
  constexpr bool operator()(pair<int, int> const& a,
                            pair<int, int> const& b) const noexcept {
    return (a.first < b.first) || (a.first == b.first && a.second < b.second);
  }
};
int main() {
  long long int n, i, t, k;
  cin >> n;
  long long int arr[2 * n];
  long long int sum = 0;
  for (long long int i = 0; i < 2 * n; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  long long int sum1 = 0;
  if (sum % 2 == 1) {
    for (long long int i = 0; i < 2 * n; i++) cout << arr[i] << " ";
  } else {
    sort(arr, arr + 2 * n);
    for (long long int i = 0; i < n; i++) {
      sum1 += arr[i];
    }
    if (sum1 == sum / 2) {
      cout << -1;
    } else {
      for (long long int i = 0; i < 2 * n; i++) cout << arr[i] << " ";
    }
  }
}
