# 奇怪的要求（Symmetric Order）

## 前言

闲着没事干，找找水题做做。疯狂玩游戏的代价，空虚寂寞冷，千万不要学我。

## 题目

Symmetric Order

In your job at Albatross Circus Management (yes, it's run by a bunch of clowns), you have just finished writing a program whose output is a list of names in nondescending order by length (so that each name is at least as long as the one preceding it). However, your boss does not like the way the output looks, and instead wants the output to appear more symmetric, with the shorter strings at the top and bottom and the longer strings in the middle. His rule is that each pair of names belongs on opposite ends of the list, and the first name in the pair is always in the top part of the list. In the first example set below, Bo and Pat are the first pair, Jean and Kevin the second pair, etc.

Input

The input consists of one or more sets of strings, followed by a final line containing only the value 0. Each set starts with a line containing an integer, n, which is the number of strings in the set, followed by n strings, one per line, sorted in nondescending order by length. None of the strings contain spaces. There is at least one and no more than 15 strings per set. Each string is at most 25 characters long.

Output

For each input set print "SET n" on a line, where n starts at 1, followed by the output set as shown in the sample output.

Sample Input

7
Bo
Pat
Jean
Kevin
Claude
William
Marybeth
6
Jim
Ben
Zoe
Joey
Frederick
Annabelle
5
John
Bill
Fran
Stan
Cece
0


Sample Output

SET 1
Bo
Jean
Claude
Marybeth
William
Kevin
Pat
SET 2
Jim
Zoe
Frederick
Annabelle
Joey
Ben
SET 3
John
Fran
Cece
Stan
Bill


### 假装题解

其实我们只需要按要求输出就好了（这谁都知道啊！），不过是按照他需要的奇怪顺序输出。分俩中情况：

• 奇数：依次顺序输出第2n+1（0<-n）个，直到上限个数，然后依次逆序输出第2n（上限-1<-n）
• 偶数：依次顺序输出第2n+1（0<-n）个，直到上限个数，然后依次逆序输出第2n（上限<-n)

### 假装源码

//
// Created by CongTsang on 2018/8/30.
//
#include <iostream>
#include <string>

using namespace std;

const int N=1024;

string Name[N];

int main()
{
int n,i,m=1;
while (cin>>n && n)
{
for (i = 1; i <= n; ++i) {
cin>>Name[i];
}
printf("SET %d\n",m++);

if (n%2 == 0)
{
for (i = 1; i <= n; i+=2)
cout<<Name[i]<<endl;
for (i = n; i > 0; i-=2)
cout<<Name[i]<<endl;
}
else {
for (i = 1; i <= n; i+=2)
cout<<Name[i]<<endl;
for (i = n-1; i > 0; i-=2)
cout<<Name[i]<<endl;
}
}
return 0;
}


### 题源

POJ这是提交入口，请尊重自己的学习。