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 int arr[500008], brr[500008], val[500008];
set<long long int> s, ans;
int mod = 998244353;
long long f = 0;
long long max(long long a, long long b) {
  if (a >= b) {
    return a;
  } else
    return b;
}
int main() {
  long long n, i, j, tmp, ans = 0, maximum = 0, minimum, cnt = 0, pre = 0,
                          c1 = 0, c2 = 0, q;
  cin >> n;
  n = 2 * n;
  vector<long long> a(n);
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < n - 1; i++) {
    if (a[i] != a[i + 1]) {
      break;
    }
  }
  if (i == n - 1) {
    cout << "-1";
    return 0;
  }
  sort(a.begin(), a.end());
  for (i = 0; i < n / 2; i++) {
    maximum += a[i];
  }
  for (i = n / 2; i < n; i++) {
    cnt += a[i];
  }
  if (cnt == maximum) {
    cout << "-1";
    return 0;
  }
  for (i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  return 0;
}
