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 int MOD = 1e9 + 7;
vector<int> v[100005];
int vis[100005];
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i < n; i++)
    if (n % i == 0) return false;
  return true;
}
int binomialCoeff(int n, int k) {
  if (k == 0 || k == n) return 1;
  return binomialCoeff(n - 1, k - 1) + binomialCoeff(n - 1, k);
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % MOD;
    b /= 2;
    a = (a * a) % MOD;
  }
  return ans;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  long long a[2 * n + 1];
  long long sum = 0;
  for (int i = 0; i < 2 * n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  sort(a, a + 2 * n);
  int idx = 0;
  int flag = 0;
  long long sum1 = 0;
  for (int i = 0; i < n; i++) {
    sum1 += a[i];
  }
  if ((sum - sum1) == sum1) {
    cout << "-1" << endl;
    return 0;
  } else {
    for (int i = 0; i < 2 * n; i++) {
      cout << a[i] << " ";
    }
  }
}
