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(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  cin >> n;
  int a1[n];
  long long sum1 = 0;
  int a2[n];
  long long sum2 = 0;
  for (int i = 0; i < n; i++) {
    cin >> a1[i];
    sum1 += a1[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> a2[i];
    sum2 += a2[i];
  }
  bool g = true;
  if (sum1 == sum2) {
    g = false;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (a1[i] != a2[j]) {
          swap(a1[i], a2[j]);
          g = true;
          i = n;
          break;
        }
      }
    }
  }
  if (g) {
    for (int i = 0; i < n; i++) {
      cout << a1[i] << ' ';
    }
    for (int i = 0; i < n - 1; i++) {
      cout << a2[i] << ' ';
    }
    cout << a2[n - 1] << '\n';
  } else {
    cout << "-1\n";
  }
  return 0;
}
