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<int> read_int_array(int n) {
  vector<int> arr;
  for (int i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    arr.push_back(temp);
  }
  return arr;
}
vector<long long int> read_llint_array(int n) {
  vector<long long int> arr;
  for (int i = 0; i < n; i++) {
    long long int temp;
    cin >> temp;
    arr.push_back(temp);
  }
  return arr;
}
bool isprime(int n) {
  if (n == 1 || n == 0) {
    return false;
  }
  if (n == 2) {
    return true;
  }
  for (int i = 2; i < n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
bool isvowel(char x) {
  if (x == 'a' || x == 'e' || x == 'o' || x == 'i' || x == 'u') {
    return true;
  } else {
    return false;
  }
}
int main() {
  int n;
  cin >> n;
  vector<int> arr = read_int_array(2 * n);
  sort(arr.begin(), arr.end());
  bool dif = false;
  int i = 1;
  while (i < 2 * n) {
    if (arr[i] != arr[i - 1]) {
      dif = true;
      break;
    }
    i++;
  }
  if (!dif) {
    cout << -1;
  } else {
    for (int i = 0; i < 2 * n; i++) {
      cout << arr[i] << " ";
    }
  }
  return 0;
}
