Codeforces Round #446 (Div. 2) D. Gluttony (构造贪心)

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本文介绍了一个算法问题，即如何构造一个独特的数组排列b，使得它与给定数组a的任意相同位置子集和都不同。文章提供了一种通过重新排列元素来解决此问题的有效方法。

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.

$\text{[math]}$

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.

#include <bits/stdc++.h>
using namespace std;
vector<pair<int,int> >a;
int b[222];
int main(){
	int n;
	cin >> n;
	int x;
	for(int i = 1; i <= n; ++i){
		cin >> x;
		a.push_back(make_pair(x, i - 1));
	}
	sort(a.begin(), a.end());
	for(int i = 0; i < n - 1; ++i){
		b[a[i].second] = a[i + 1].first;
	}
	b[a[n - 1].second] = a[0].first;
	for(int i = 0; i < n; ++i){
		cout<<b[i]<<" ";
	}
}

/*
题意：
22个不同的数，构造一个不同的排列，使两个排列任意相同位置的数字集合和不同。

思路：
构造题，，，说实话我比赛时是猜的结论。。将最大移到次大，次大移到次次大，最小移到最大，结束。。
*/