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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct deb {
  template <class c>
  deb &operator<<(const c &) {
    return *this;
  }
};
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
const long long dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const long long dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const long long mod = 1e9 + 7;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (auto &v : vec) is >> v;
  return is;
}
const double PI = acos(-1.0);
const long long inf = 1e18;
const long long maxn = 1e3 + 5;
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr(2 * n);
  cin >> arr;
  cin >> arr;
  sort(arr.begin(), arr.end());
  long long s1 = 0;
  long long s2 = 0;
  for (long long i = 0; i < n; i++) s1 += arr[i];
  for (long long i = n; i < 2 * n; i++) s2 += arr[i];
  if (s1 != s2) {
    for (long long i = 0; i < 2 * n; i++) {
      cout << arr[i] << " ";
    }
    cout << "\n";
  } else {
    cout << -1 << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long tt = 1; tt <= t; tt++) {
    solve();
  }
}
