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 long long N = 1e5 + 5;
const long long M = 1e9 + 7;
long long power(long long a, long long b) {
  long long res = 1;
  a %= M;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % M;
    a = a * a % M;
  }
  return res;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n * 2);
  for (long long i = 0; i < 2 * n; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end());
  long long check = n;
  long long sum = 0;
  long long check1 = 0;
  for (long long i = 0; i < n * 2; i++) {
    if (i < check) {
      sum += v[i];
    } else {
      check1 += v[i];
    }
  }
  if (check1 == sum) {
    cout << -1;
  } else {
    for (long long i = 0; i < n * 2; i++) {
      cout << v[i] << " ";
    }
  }
}
int32_t main() { solve(); }
