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() {
  std::ios::sync_with_stdio(false);
  long long x, y, i, j;
  while (cin >> x) {
    vector<long long> v;
    v.clear();
    for (i = 0; i < 2 * x; i++) {
      cin >> y;
      v.push_back(y);
    }
    sort(v.begin(), v.end());
    long long sum_n1 = 0, sum_n2 = 0;
    for (i = 0; i < 2 * x; i++) {
      if (i < x) {
        sum_n1 += v[i];
      }
      if (i >= x) {
        sum_n2 += v[i];
      }
    }
    if (sum_n1 != sum_n2) {
      for (i = 0; i < 2 * x; i++) {
        cout << v[i] << " ";
      }
      cout << endl;
    } else {
      cout << "-1" << endl;
    }
  }
}
