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 int> read(long long int n) {
  vector<long long int> a;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    a.push_back(x);
  }
  return a;
}
template <typename T>
void showArr(T a) {
  for (auto i : a) cout << i << " ";
  cout << endl;
}
int main() {
  long long int n;
  cin >> n;
  vector<long long int> a = read(2 * n);
  sort(a.begin(), a.end());
  reverse(a.begin(), a.end());
  long long int sum = 0;
  for (auto x : a) {
    sum += x;
  }
  long long int sumx = 0;
  for (long long int i = 0; i < a.size() / 2; i++) {
    sumx += a[i];
  }
  if (sum - sumx == sumx) {
    cout << -1 << endl;
  } else {
    showArr(a);
  }
  return 0;
}
