Codeforce 633C. Spy Syndrome 2

C. Spy Syndrome 2

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and ancient as Caesar cipher. After many weeks of observation of Siddhant’s sentences, Yash determined a new cipher technique.

For a given sentence, the cipher is processed as:

1. Convert all letters of the sentence to lowercase.
2. Reverse each of the words of the sentence individually.
3. Remove all the spaces in the sentence.

For example, when this cipher is applied to the sentence

Kira is childish and he hates losing

the resulting string is

ariksihsidlihcdnaehsetahgnisol

Now Yash is given some ciphered string and a list of words. Help him to find out any original sentence composed using only words from the list. Note, that any of the given words could be used in the sentence multiple times.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 10 000) — the length of the ciphered text. The second line consists of nlowercase English letters — the ciphered text t.

The third line contains a single integer m (1 ≤ m ≤ 100 000) — the number of words which will be considered while deciphering the text. Each of the next m lines contains a non-empty word wi (|wi| ≤ 1 000) consisting of uppercase and lowercase English letters only. It's guaranteed that the total length of all words doesn't exceed 1 000 000.

Output

Print one line — the original sentence. It is guaranteed that at least one solution exists. If there are multiple solutions, you may output any of those.

Examples

input

30
ariksihsidlihcdnaehsetahgnisol
10
Kira
hates
is
he
losing
death
childish
L
and
Note

output

Kira is childish and he hates losing 

input

12
iherehtolleh
5
HI
Ho
there
HeLLo
hello

output

HI there HeLLo 

Note

In sample case 2 there may be multiple accepted outputs, "HI there HeLLo" and "HI there hello" you may output any of them.

思路：暴力，字典树+DFS。感觉这题有点水过去的感觉。

说下我的思路，就是将下面给的单词到过来存进字典树中，然后用DFS正着搜。时间给了2s，140ms过的，也差点爆了内存

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<math.h>
  6 #include<queue>
  7 #include<map>
  8 using namespace std;
  9 typedef long long LL;
 10 char str[10005];
 11 char trr[10005];
 12 char strr[1005];
 13 char tt[100005][1001];
 14 int BB;
 15 struct node
 16 {
 17     node*p[26];
 18     int id;//记录单词的标号
 19     node()
 20     {
 21         id=-2;
 22         memset(p,0,sizeof(p));
 23     }
 24 };
 25 int N=0;
 26 int ii[100005];
 27 int rr[100005];
 28 node *head=new node();
 29 void inserts(struct node*r,char *q,int v)//字典树的添加
 30 {
 31     int i,j,l;
 32     node*NN=r;
 33     l=strlen(q);
 34     for(i=l-1; i>=0; i--)
 35     {
 36         int s=tolower(q[i])-'a';
 37         if(NN->p[s]==NULL)
 38         {
 39             node*n=new node();
 40             NN->p[s]=n;
 41             NN=n;
 42         }
 43         else
 44             NN=NN->p[s];
 45         if(i==0)
 46             NN->id=v;
 47     }
 48 }
 49 struct node*ask(char p,struct node*r)//查询，返回当前的节点
 50 {
 51     int i,j;
 52     struct node*dd=r;
 53         int s=p-'a';
 54         return dd->p[s];
 55 }
 56 void dfs(int id,int l,int k,node*d)
 57 {
 58     if(N)return ;
 59     int i;
 60     char tr[10005];
 61     if(id==l&&N==0)
 62     {for(i=0;;i++)
 63     {
 64         if(ii[i])
 65         {
 66             rr[i]=ii[i];
 67         }
 68         else break;
 69     }
 70         N=1;
 71         return ;
 72     }node *WW;
 73     for(i=id; i<l; i++)
 74     {
 75         if(N)return ;
 76         tr[i-id]=str[i];
 77         tr[i-id+1]='\0';
 78         if(i==id)WW=head;
 79         node *nn=ask(tr[i-id],WW);
 80         if(nn==NULL)//如果没有符合的直接跳出
 81         {
 82             return ;
 83         }
 84         if(nn->id!=-2)//当符合了进入下一层
 85         {
 86             ii[k]=nn->id;
 87             dfs(i+1,l,k+1,nn);
 88             ii[k]=0;
 89         } WW=nn;//继续按当前的前缀串查找
 90     }
 91 }
 92 int main(void)
 93 {
 94     int i,j,k,s;
 95     N=0,BB=0;
 96     scanf("%d",&k);
 97     scanf("%s",str);
 98     scanf("%d",&s);
 99     int ans=0;
100     for(i=1; i<=s; i++)
101     {
102         scanf("%s",tt[i]);
103         int l=strlen(tt[i]);
104         inserts(head,tt[i],i);
105     }
106     dfs(0,k,0,head);
107     for(i=0;;i++)
108     {
109         if(rr[i]==0)
110         {
111             break;
112         }
113     }
114     int sl=i-1;
115     printf("%s",tt[rr[0]]);
116     for(i=1; i<=sl; i++)
117     {
118         printf(" %s",tt[rr[i]]);
119     }
120     return 0;
121 }