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 sum1, sum2;
bool swa(vector<int>& a, int n) {
  int aux;
  for (size_t i = 0; i < a.size() / 2; i++) {
    if (a[i] != a[i + n]) {
      aux = a[i];
      a[i] = a[i + n];
      a[i + n] = aux;
      return true;
    } else {
      for (size_t j = 1; j < n; j++) {
        if (a[i] != a[j + n]) {
          aux = a[i];
          a[i] = a[j + n];
          a[j + n] = aux;
          return true;
        }
      }
    }
  }
  return false;
}
int main(int argc, char const* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a;
  bool ans;
  for (size_t i = 0; i < 2 * n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
    if (i < n)
      sum1 += x;
    else
      sum2 += x;
  }
  if (sum1 != sum2) {
    for (auto i : a) cout << i << " ";
    return 0;
  } else {
    ans = swa(a, n);
  }
  if (ans)
    for (auto i : a) cout << i << " ";
  else
    cout << -1 << endl;
  return 0;
}
