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;
long long fac[2000005] = {0};
void factorial() {
  fac[0] = 1;
  fac[1] = 1;
  for (long long i = 2; i < 200005; i++)
    fac[i] = ((i % 1000000007) * (fac[i - 1] % 1000000007)) % 1000000007;
}
long long power(long long n, long long m) {
  long long p = 1;
  if (m == 0) return 1;
  p = (power(n, m / 2) % 1000000007);
  p = (p % 1000000007 * p % 1000000007) % 1000000007;
  return (m & 1 ? ((p % 1000000007 * n % 1000000007) % 1000000007)
                : (p % 1000000007));
}
long long ncr(long long n, long long r) {
  return ((fac[n] * power(fac[r], 1000000007 - 2)) % 1000000007 *
          power(fac[n - r], 1000000007 - 2)) %
         1000000007;
}
void ram5564() {
  long long n;
  cin >> n;
  n *= 2;
  set<long long> s;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s.insert(a[i]);
  }
  if ((long long)s.size() == 1) {
    cout << -1;
    return;
  }
  sort(a, a + n);
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) ram5564();
  return 0;
}
