Codeforces Round #201 (Div. 2) 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 sequencea. 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

5100 -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.

#include <iostream>#include <stdio.h>#include <algorithm>using namespace std;#define M 10500bool cmp(int a,int b){    return a<b;}int pri[M];int main(){   int n,i;   while(scanf("%d",&n)!=EOF){      for(i=0;i<n;i++){        scanf("%d",&pri[i]);      }      sort(pri,pri+n,cmp);      printf("%d",pri[n-1]);      for(i=1;i<n-1;i++){        printf(" %d",pri[i]);      }      printf(" %d\n",pri[0]);   }    return 0;}