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.
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.

一开始让flag数组记录每个数的位置，然后从小到大排序，排序后的flag是指a中数从小到大的数所对应的位置。然后让前n-1个a中的数的位置分别在b中对应比它小一点的数，最后让a中最小的数的位置在b中为a的最大的数。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
int a[25], b[25], flag[25];
int cmp(int x, int y)
{
  return a[x]<a[y];
}
int main()
{
  int n;
  scanf("%d", &n);
  for (int i=1; i<=n; i++){
    scanf("%d", &a[i]);
    flag[i]=i;
  }
  sort(flag+1, flag+n+1, cmp);
  for (int i=1; i<n; i++){
    b[flag[i+1]]=a[flag[i]];
  }
  b[flag[1]]=a[flag[n]];
  for (int i=1; i<=n; i++){
    printf("%d", b[i]);
    if (i!=n)
      printf(" ");
  }
  return 0;
}