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;
  int a[2 * n];
  int ans_min = 0;
  int ans_max = 10000001;
  long s1 = 0, s2 = 0;
  int ind_max, ind_min;
  int ma1 = ans_min, ma2 = ans_min, mi1 = ans_max, mi2 = ans_max;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ma1 = max(ma1, a[i]);
    if (ma1 == a[i]) {
      ind_max = i;
    }
    mi1 = min(mi1, a[i]);
    s1 = s1 + a[i];
  }
  for (int i = n; i < 2 * n; i++) {
    cin >> a[i];
    ma2 = max(ma2, a[i]);
    mi2 = min(mi2, a[i]);
    if (mi2 == a[i]) {
      ind_min = i;
    }
    s2 = s2 + a[i];
  }
  if ((ma1 == mi1) && (ma2 == mi2) && (ma1 == ma2)) {
    cout << "-1";
  } else if (s1 == s2) {
    a[ind_max] = mi2;
    a[ind_min] = ma1;
    for (int i = 0; i < 2 * n; i++) {
      cout << a[i] << " ";
    }
  } else {
    for (int i = 0; i < 2 * n; i++) {
      cout << a[i] << " ";
    }
  }
}
