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;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long seve[1000000] = {0};
set<long long> check;
void seive() {
  long long i, j;
  seve[0] = 1;
  seve[1] = 1;
  for (i = 2; i <= 1000000; i++) {
    if (seve[i] == 0) {
      check.insert(i);
      for (j = i * i; j <= 1000000; j = j + i) seve[j] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long m, i, j = 0, f = 0, k = 0, y, x, n;
  string s;
  cin >> n;
  long long a[2 * n];
  for (i = 0; i < 2 * n; i++) cin >> a[i];
  sort(a, a + 2 * n);
  if (a[0] == a[2 * n - 1])
    cout << "-1";
  else
    for (i = 0; i < 2 * n; i++) {
      cout << a[i] << " ";
    }
}
