#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
stringstream output;
inline void eOP();
inline void no() {
  output.str(string());
  output << -1;
  eOP();
}
int main() {
  int n;
  cin >> n;
  set<int> nums;
  int* a = new int[n];
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (nums.count(a[i])) {
      no();
    }
    nums.insert(a[i]);
  }
  vector<int> b;
  for (auto it = (nums).begin(); it != (nums).end(); ++it) {
    b.push_back(*it);
  }
  map<int, int> m;
  for (int i = 0; i < n; ++i) {
    m[b[i]] = b[(i + 1) % n];
  }
  for (int i = 0; i < n; ++i) {
    output << m[a[i]] << " ";
  }
  eOP();
}
inline void eOP() {
  cout << output.str();
  output.str(string());
  exit(0);
}

### Prompt

Construct a CPP code solution to the problem outlined:

You are given an array a with n distinct integers. Construct an array b by permuting a such that for every non-empty subset of indices S = {x1, x2, ..., xk} (1 ≤ xi ≤ n, 0 < k < n) the sums of elements on that positions in a and b are different, i. e. 

<image>

Input

The first line contains one integer n (1 ≤ n ≤ 22) — the size of the array.

The second line contains n space-separated distinct integers a1, a2, ..., an (0 ≤ ai ≤ 109) — the elements of the array.

Output

If there is no such array b, print -1.

Otherwise in the only line print n space-separated integers b1, b2, ..., bn. Note that b must be a permutation of a.

If there are multiple answers, print any of them.

Examples

Input

2
1 2


Output

2 1 


Input

4
1000 100 10 1


Output

100 1 1000 10

Note

An array x is a permutation of y, if we can shuffle elements of y such that it will coincide with x.

Note that the empty subset and the subset containing all indices are not counted.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
stringstream output;
inline void eOP();
inline void no() {
  output.str(string());
  output << -1;
  eOP();
}
int main() {
  int n;
  cin >> n;
  set<int> nums;
  int* a = new int[n];
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (nums.count(a[i])) {
      no();
    }
    nums.insert(a[i]);
  }
  vector<int> b;
  for (auto it = (nums).begin(); it != (nums).end(); ++it) {
    b.push_back(*it);
  }
  map<int, int> m;
  for (int i = 0; i < n; ++i) {
    m[b[i]] = b[(i + 1) % n];
  }
  for (int i = 0; i < n; ++i) {
    output << m[a[i]] << " ";
  }
  eOP();
}
inline void eOP() {
  cout << output.str();
  output.str(string());
  exit(0);
}

```