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 int mod = 1e9 + 7;
const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
void solve(int test_case) {
  int n;
  cin >> n;
  vector<int> v(n + n);
  for (int i = 0; i < n + n; i++) cin >> v[i];
  sort((v).begin(), (v).end());
  long long s1 = 0;
  long long s2 = 0;
  for (int i = 0; i < n; i++) {
    s1 += v[i];
    s2 += v[i + n];
  }
  if (s1 == s2)
    cout << -1;
  else {
    for (auto &val : v) cout << val << " ";
  }
  cout << "\n";
  ;
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}
