codeforces Round #446 (Div. 2) D

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

21 2

output

2 1 

input

41000 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.

题意：给你一个数列没有重复的数，问你怎么交换使得

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.

• 题解：将每一个数换成后一个比它大的数，最大的数换成最小的那一个，就行了。（或者每个数换成前一个比小的之类多可以）

考虑分两种情况讨论：
1.原子集没有最大的那个数：这就显然是个正确的，每个数都比之前的数要大，所以总和肯定要大；

2.原子集有最大的那个数：最大数改变后的值最整个里面最大的，然后你将这个子集去掉最大的那个数，剩下的变化量的和肯定要小于你最大值变为最小值的值，因为你之间肯定会有多余的空格（因为不能为全集）留着，所以肯定是可行的。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>using namespace std;#define INF 0x3f3f3f3f#define N 2010struct node {    int val;    int pos;}a[N];int b[N];bool cmp(node a, node b){    if(a.val == b.val)        return a.pos < b.pos;    return a.val < b.val;}int main(){    int n;    scanf("%d", &n);    for(int i = 1; i <=  n; i++) {        cin >> a[i].val;        a[i].pos = i;    }    sort(a + 1, a + n + 1, cmp);    for(int i = 1; i <= n; i++)        b[a[i].pos] = i;    for(int i = 1; i <= n; i++) {        if(b[i] != n)            cout << a[b[i] + 1].val << " ";        else            cout << a[1].val << " ";    }    return 0;}