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;
template <class T>
void printQueue(T& q) {
  while (!q.empty()) {
    cout << q.top() << endl;
    q.pop();
  }
}
long long power(long long a, long long b) {
  long long r = 1;
  while (b != 0) {
    if (b % 2 != 0) {
      r = r * a;
    }
    a = a * a;
    b = b / 2;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long ast[2 * n + 1];
    for (int i = 1; i <= (2 * n); i++) {
      cin >> ast[i];
    }
    sort(ast + 1, ast + n + n + 1);
    long long sum = 0;
    for (int i = 1; i <= n; i++) {
      sum += ast[i];
    }
    long long sum2 = 0;
    for (int i = n + 1; i <= (2 * n); i++) {
      sum2 += ast[i];
    }
    if (sum == sum2) {
      cout << -1 << endl;
    } else {
      for (int i = 1; i <= 2 * n; i++) {
        cout << ast[i] << " ";
      }
      cout << endl;
    }
  }
}
