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.

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.

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.

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.

题目大意：给定一个数组a,将a中的元素调换位置变成b数组，使得对于每一个下标子集，a中对应下标的元素和不等于b中对应下标的元素和.

分析：构造题.如果第i位的数是第j大的，那么就把第j+1大的放在第i位，如果已经是最大的了，就把最小的放在第i位，这样一定是对的.证明的话就是看是否选最大的那个数.

这个数据范围真的是让人分分钟想到状压......

#include <cstdio>

#include <cmath>

#include <queue>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

int n, a[];

struct node

{

    int x, id;

}e[];

bool cmp(node a, node b)

{

    return a.x < b.x;

}

int main()

{

    scanf("%d", &n);

    for (int i = ; i <= n; i++)

    {

        scanf("%d", &e[i].x);

        e[i].id = i;

    }

    sort(e + , e +  + n, cmp);

    for (int i = ; i <= n; i++)

        a[e[i].id] = i;

    for (int i = ; i <= n; i++)

    {

        int t = (a[i] + ) % n;

        if (t == )

            t = n;

        printf("%d ", e[t].x);

    }

    return ;

}