A - Difference Row

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探讨如何求解特定序列排列的问题，旨在找到使相邻元素差值之和最大化的排列方式，并确保该排列是最小字典序。通过分析得出结论，最大值应置于首位，最小值置于末位，其余元素按升序排列。

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：

就是找出(x₁ - x₂) + (x₂ - x₃) + ... + (x_n - 1 - x_n)的最大值，并输出最小的排序

分析：

去掉（）就是第一个减最后一个，所以最大值放在第一个，最小值放在最后，中间从小到大排列

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,i,a[200];
    while(cin>>n)
    {
        for(i=0;i<n;i++)
            cin>>a[i];
        sort(a,a+n);
        swap(a[0],a[n-1]);
        for(i=0;i<n-1;i++)
            cout<<a[i]<<" ";
        cout<<a[n-1]<<endl;
    }
}

感受：

sort排序真方便