Codeforces 347A A. Difference Row

A. Difference Row

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You want to arrange n integers a1, a2, ..., an 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, ..., xn, where sequence x is a permutation of sequence a. The value of such an arrangement is (x1 - x2) + (x2 - x3) + ... + (xn - 1 - xn).

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, ..., an (|ai| ≤ 1000).

Output

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

Sample test(s)

input

5
100 -100 50 0 -50

output

100 -50 0 50 -100 

Note

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, ... , xp is lexicographically smaller than sequence y1, y2, ... , yp if there exists an integer r (0 ≤ r < p) such that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1.

思路：好水

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int map[110];
int n;
int main()
{
while(~scanf("%d",&n))
{
memset(map,0,sizeof(map));
for(int i = 0;i < n;i ++)
scanf("%d",&map[i]);
sort(map,map + n);
printf("%d ",map[n - 1]);
for(int i = 1;i < n - 1;i ++)
printf("%d ",map[i]);
printf("%d ",map[0]);
}
return 0;
}