Codeforces 892 D.Gluttony

D. Gluttony

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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 ;
}