Codeforces Round #446 (Div. 2) D. Gluttony (構造 貪心)

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.

#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]<<" ";
	}
}

/*
題意：
22個不同的數，構造一個不同的排列，使兩個排列任意相同位置的數字集合和不同。

思路：
構造題，，，說實話我比賽時是猜的結論。。将最大移到次大，次大移到次次大，最小移到最大，結束。。
*/