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;
int main() {
  int n;
  cin >> n;
  vector<int> v1, v2;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    v1.push_back(p);
  }
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    v2.push_back(p);
  }
  int t = 0;
  for (int i = 0; i < n; i++) {
    int sum1 = 0, sum2 = 0;
    for (int j = 0; j < n; j++) {
      if (i == 0 && t == 0) {
        sum1 += v1[j];
        sum2 += v2[j];
      } else if (sum1 == sum2) {
        if (v1[i] != v2[j]) {
          int c = v1[i];
          v1[i] = v2[j];
          v2[j] = c;
          sum1 = sum1 - v1[i] + v2[j];
          sum2 = sum2 - v2[j] + v1[i];
          break;
        }
      }
    }
    if (i == 1 && t == 0) {
      i = 0;
      t = 5;
    }
    if (sum1 != sum2) {
      break;
    }
    if (i == n - 1) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << v1[i] << " ";
  }
  for (int i = 0; i < n; i++) {
    cout << v2[i] << " ";
  }
  return 0;
}
