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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    n *= 2;
    vector<long long> arr;
    int x;
    for (int i = 0; i < (int)(n); ++i) {
      cin >> x;
      arr.push_back((long long)x);
    }
    sort(arr.begin(), arr.end());
    long long sumb = 0, sume = 0;
    for (int i = 0; i < (int)(n / 2); ++i) {
      sumb += arr[i];
      sume += arr[n - i - 1];
    }
    if (sume != sumb) {
      for (auto it = arr.begin(); it != arr.end(); it++) {
        cout << *it << " ";
      }
    } else {
      cout << "-1";
    }
  }
}
