Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

You want to arrange n integers a₁,?a₂,?...,?a_n in some order in a row. Let‘s define the value of an arrangement as the sum of differences between all pairs of adjacent integers.

More formally, let‘s denote some arrangement as a sequence of integers x₁,?x₂,?...,?x_n, where sequence x is a permutation of sequence a. The value of such an arrangement is (x₁?-?x₂)?+?(x₂?-?x₃)?+?...?+?(x_n?-?1?-?x_n).

Find the largest possible value of an arrangement. Then, output the lexicographically smallest sequence x that corresponds to an arrangement of the largest possible value.

Input

The first line of the input contains integer n (2?≤?n?≤?100). The second line contains n space-separated integers a₁, a₂, ..., a_n (|a_i|?≤?1000).

Output

Print the required sequence x₁,?x₂,?...,?x_n. Sequence x should be the lexicographically smallest permutation of a that corresponds to an arrangement of the largest possible value.

Sample Input

Input

5
100 -100 50 0 -50

Output

100 -50 0 50 -100 

Hint

In the sample test case, the value of the output arrangement is (100?-?(?-?50))?+?((?-?50)?-?0)?+?(0?-?50)?+?(50?-?(?-?100))?=?200. No other arrangement has a larger value, and among all arrangements with the value of 200, the output arrangement is the lexicographically smallest one.

Sequence x₁,?x₂,?... ,?x_p is lexicographically smaller than sequence y₁,?y₂,?... ,?y_p if there exists an integer r(0?≤?r?<?p) such that x₁?=?y₁,?x₂?=?y₂,?... ,?x_r?=?y_r and x_r?+?1?<?y_r?+?1.

题意：

给定一个序列，求使得(x₁?-?x₂)?+?(x₂?-?x₃)?+?...?+?(x_n?-?1?-?x_n)最大且字典序最小的排列

化简式子，得到：x1-xn  即求出x1-xn最大的即可

然后按字典序最小的输出

代码：

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
int a[130];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    int x,y,res=-INF;
    int xl,yl;
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            if(a[i]-a[j]>res){
                x=a[i],y=a[j];
                xl=i,yl=j;
                res=a[i]-a[j];
            }
            if(a[j]-a[i]>res){
                y=a[i],x=a[j];
                xl=j,yl=i;
                res=a[j]-a[i];
            }
        }
    }
    a[xl]=-INF,a[yl]=-INF;
    sort(a,a+n);
    cout<<x<<" ";
    for(int i=0;i<n;i++){
        if(a[i]!=-INF){
            cout<<a[i]<<" ";
        }
    }
    cout<<y<<endl;
}

codefroces Round #201.a--Difference Row