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;
vector<long long> g[7];
map<int, int> pos1, pos2;
const int MAX = 2e5 + 5;
long long ara[MAX];
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = binpow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, k, x, s1 = 0, s2 = 0;
  cin >> n;
  n = n * 2;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (long long i = 0; i < n / 2; i++) s1 += a[i];
  for (long long i = n / 2; i < n; i++) s2 += a[i];
  if (s1 == s2)
    cout << -1;
  else {
    for (long long i = 0; i < n; i++) cout << a[i] << " ";
  }
  return 0;
}
