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;
long long GCD(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0)
    return a;
  else
    return GCD(b, a % b);
}
int32_t main() {
  {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  };
  long long n;
  cin >> n;
  vector<long long> v1, v2;
  long long sum1 = 0, sum2 = 0;
  long long min = INT_MAX, mini = -1;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    sum1 += x;
    if (min > x) min = x, mini = i;
    v1.push_back(x);
  }
  long long Max = INT_MIN, maxi = -1;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    sum2 += x;
    if (Max < x) Max = x, maxi = i;
    v2.push_back(x);
  }
  if (sum1 != sum2) {
    for (long long i = 0; i < n; i++) cout << v1[i] << " ";
    for (long long i = (0); i <= ((long long)(n)-1); i++) {
      cout << v2[i] << " ";
    }
  } else {
    if (min == Max) {
      cout << -1 << endl;
      return 0;
    }
    v1[mini] = Max;
    v2[maxi] = min;
    for (long long i = 0; i < n; i++) cout << v1[i] << " ";
    for (long long i = (0); i <= ((long long)(n)-1); i++) {
      cout << v2[i] << " ";
    }
  }
}
