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 target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 998244353;
double eps = 1e-12;
const int N = 300500;
const int mod = 1000000007;
int nCr(int n, int k) {
  int C[k + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1];
  }
  return C[k];
}
bool isdis(string s) {
  set<char> st(s.begin(), s.end());
  return (st.size() == s.length());
}
void solve() {
  long long k;
  cin >> k;
  long long n = 2 * k;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long sum1 = 0, sum2 = 0;
  for (long long i = 0; i < n / 2; i++) sum1 += a[i];
  for (long long i = n / 2; i < n; i++) sum2 += a[i];
  if (sum1 == sum2) {
    cout << "-1\n";
    return;
  }
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}
