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 b2(const pair<pair<int, int>, int> &p1,
        const pair<pair<int, int>, int> &p2) {
  return p1.first.first > p2.first.first;
}
bool b1(const pair<pair<int, int>, int> &p1,
        const pair<pair<int, int>, int> &p2) {
  return p1.first.second < p2.first.second;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n;
  cin >> n;
  n = n * 2;
  long long int sum = 0, sum1 = 0;
  int arr[n];
  for (int i = 0; i <= n - 1; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  sort(arr, arr + n);
  for (int i = 0; i < n / 2; i++) sum1 += arr[i];
  if (sum1 * 2 == sum)
    cout << "-1" << endl;
  else {
    for (int i = 0; i <= n - 1; i++) cout << arr[i] << " ";
    cout << endl;
  }
  return 0;
}
