#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, sum = 0, sum1 = 0, sum2 = 0;
  cin >> n;
  int m[2 * n];
  for (int i = 0; i < n * 2; i++) {
    cin >> m[i];
    sum = sum + m[i];
  }
  for (int i = 0; i < n; i++) {
    sum1 = sum1 + m[i];
    sum2 = sum2 + m[i + n];
  }
  if (sum1 != sum2) {
    for (int i = 0; i < n * 2; i++) {
      cout << m[i] << ' ';
    }
  } else {
    sort(m, m + n * 2);
    for (int i = 0; i < n; i++) {
      sum1 = sum1 + m[i];
      sum2 = sum2 + m[i + n];
    }
    if (sum1 != sum2) {
      for (int i = 0; i < n * 2; i++) {
        cout << m[i] << ' ';
      }
    } else {
      cout << "-1";
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, sum = 0, sum1 = 0, sum2 = 0;
  cin >> n;
  int m[2 * n];
  for (int i = 0; i < n * 2; i++) {
    cin >> m[i];
    sum = sum + m[i];
  }
  for (int i = 0; i < n; i++) {
    sum1 = sum1 + m[i];
    sum2 = sum2 + m[i + n];
  }
  if (sum1 != sum2) {
    for (int i = 0; i < n * 2; i++) {
      cout << m[i] << ' ';
    }
  } else {
    sort(m, m + n * 2);
    for (int i = 0; i < n; i++) {
      sum1 = sum1 + m[i];
      sum2 = sum2 + m[i + n];
    }
    if (sum1 != sum2) {
      for (int i = 0; i < n * 2; i++) {
        cout << m[i] << ' ';
      }
    } else {
      cout << "-1";
    }
  }
  return 0;
}

```