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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long N = 998244353, M = 1000000;
long long n, m, t, k, k2, f, mi = 2e9, x, y, z, ma, x3, x2, y2, l, r, i, j, ans,
                             a[410000], b[410000], d[410000], dp[410000],
                             p[410000];
string s, s2;
vector<long long> v, v2, v3;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  n *= 2;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (i = 1; i <= n / 2; i++) {
    x += a[i];
  }
  for (i = n / 2 + 1; i <= n; i++) {
    y += a[i];
  }
  if (x == y)
    cout << "-1";
  else
    for (i = 1; i <= n; i++) cout << a[i] << ' ';
}
