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;
bool compare(pair<int, double> a, pair<int, double> b) {
  if (a.second > b.second)
    return 1;
  else
    return 0;
}
struct monster {
  double a, b, prof;
  int ind;
};
bool compa(monster a, monster b) {
  if (a.prof > b.prof)
    return 1;
  else
    return 0;
}
bool com(monster a, monster b) {
  if (a.ind < b.ind)
    return 1;
  else
    return 0;
}
void solve() {
  long long n;
  cin >> n;
  long long l = 2 * n;
  long long arr[l];
  for (long long i = 0; i < l; i++) cin >> arr[i];
  sort(arr, arr + l);
  if (accumulate(arr, arr + l, 0) == arr[0] * l) {
    cout << -1 << "\n";
    return;
  }
  for (long long i = 0; i < l; i++) cout << arr[i] << " ";
  cout << "\n";
}
int main() { solve(); }
