codeforces 652B z-sort（思维）

B. z-sort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:

1. ai ≥ ai - 1 for all even i,
2. ai ≤ ai - 1 for all odd i > 1.

For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’tz-sorted.

Can you make the array z-sorted?

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.

The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.

Output

If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".

Examples

input

4
1 2 2 1

output

1 2 1 2

input

5
1 3 2 2 5

output

1 5 2 3 2

题意：给你一组数问是否可以进行Z排序    z排序要求：对于奇数位i 满足ai<=ai-1 偶数位i满足ai>=ai-1  可以的话输出排序后的解
题解：因为有等号，所以就简单多了，我们将原数组从小到大排序，然后将后一半大的数放在偶数位即可，显然所有的序列都满足

#include<stdio.h>
#include<string.h>
#include<queue>
#include<cstdio>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 6000
#define mod 100
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int s[MAX];
int a[MAX];
int ans[MAX],op[MAX];
int main()
{
	int n,m,j,i,t,k,o,l;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			//a[i]=s[i];
		}
		sort(a+1,a+n+1);
		m=n/2;
		if(n&1) m=m+2;
		else m=m+1;
		j=1;
		for(i=n;i>=m;i--)
			op[j++]=a[i];
		for(i=1;i<m;i++)
		    s[i]=a[i];
			//printf("%d* ",a[i]);
		//printf("\n");
		l=o=1;
		for(i=1;i<=n;i++)
		{
			if(i&1) ans[i]=s[l++];
			else ans[i]=op[o++];
		}
		for(i=1;i<=n;i++)
		printf("%d ",ans[i]);
		printf("\n");
	}
	return 0;
}