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 double PI = 3.1415926535897923846;
const long long int MOD = 1e9 + 7;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(2 * n);
  long long int sum = 0;
  map<long long int, long long int> mp1;
  for (long long int i = 0; i < 2 * n; i++) {
    cin >> a[i];
    mp1[a[i]]++;
  }
  sort(a.begin(), a.end());
  if (mp1.size() == 1) {
    cout << -1;
    return;
  }
  for (long long int i = 0; i < 2 * n; i++) {
    cout << a[i] << " ";
  }
}
int main() {
  long long int test = 1;
  while (test--) solve();
}
