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;
const int INF = 0x3f3f3f3f;
const long long LLINF = 1e18;
const int maxN = 1e5;
unordered_map<int, int> mp;
int arr[2001];
int main() {
  int n;
  scanf("%d", &n);
  ;
  long long sum1 = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i]);
    ;
    sum1 += arr[i];
  }
  long long sum2 = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[n + i]);
    ;
    sum2 += arr[n + i];
  }
  if (sum1 != sum2) {
    for (int i = 0; i < 2 * n; ++i) {
      printf("%d ", arr[i]);
    }
  } else {
    for (int i = 0; i < n; ++i) {
      if (mp.find(arr[i]) != mp.end()) continue;
      mp[arr[i]] = 1;
      for (int j = 0; j < n; ++j) {
        if (arr[i] != arr[n + j]) {
          int aux = arr[i];
          arr[i] = arr[n + j];
          arr[n + j] = aux;
          for (int k = 0; k < 2 * n; ++k) {
            printf("%d ", arr[k]);
          }
          printf("\n");
          return 0;
        }
      }
    }
    printf("-1\n");
  }
  return 0;
}
