FJ is about to take his?N?(1 ≤?N?≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges his cows in a line and herds them past the judges.

The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows' names.

FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.

FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he's finished, FJ takes his cows for registration in this new order.

Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.

Input

* Line 1: A single integer:?N
* Lines 2..N+1: Line?i+1 contains a single initial ('A'..'Z') of the cow in the?ith position in the original line

Output

The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows ('A'..'Z') in the new line.

Sample Input

6 A C D B C B

Sample Output

ABCBCD

*這題主要是用貪心，目標主要是從已知的字符串構造出字典序盡可能小的字符串
*從字典序性質來看，無論字符串末尾有多大，只要保證前面部分較小就可以咯！
*假設原來字符串為S，目標字符串為T，那么，不斷取出S的開頭和末尾較小的一個
*字符放到T的末尾。上面的沒有針對開頭與結尾相同，如果，遇到這樣動情況，應
*該比較下一個字符的大小。如果，下一個字符也相同，那么可以得到下面的算法：
*按照字典序比較S和將S反轉后的字符串S’
*如果，S較小，那么取出S開頭字符，追加到T的末尾
*如果，S'較小，那么取出S結尾字符，追加到T的末尾
*如果，相同則取出那一個都可以。

?

#include <iostream> #include <stdio.h> using namespace std; int n; char s[2100];int main() {scanf("%d",&n);int cnt=0;char temp[2];while(cnt<n){scanf("%s",temp);s[cnt++]=temp[0];}int f=0,e=n-1,cnt0=0;while(f <= e) // 將從左起和從右起的字符串比較{bool flag = false;for(int i = 0; i <= e - f; i++) // 字符串的開頭與結尾相同，則繼續(xù)比較下一個{if(s[f+i] < s[e-i]){flag = true;cnt0++;break;}else if(s[f+i] > s[e-i]){flag = false;cnt0++;break;}}if(flag) cout<<s[f++];else cout<<s[e--];if(cnt0 % 80 == 0)cout<<endl;}cout << endl;//cout << "Hello world!" << endl;return 0; }

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