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:

Convert all letters of the sentence to lowercase.

Reverse each of the words of the sentence individually.

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 n lowercase 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.

字典树的应用，以单词建树，将密码串反向记忆化搜索

#include <cstdio>

#include <cstring>

#include <cmath>

#include <cstdlib>

#include <queue>

#include <stack>

#include <iostream>

#include <algorithm>

using namespace std;

const int Max = 1e6+10;

const int MaxM = 10010;

typedef struct node

{

    int next[30];

    int mark;

}Tree;

Tree Tr[Max];

int top;

char str[Max];

char s[MaxM*10][1011];

int  n,m;

int vis[MaxM];

bool flag;

int NewNode()

{

    for(int i=0;i<=26;i++)

    {

        Tr[top].next[i] = -1;

    }

    Tr[top].mark = -1;

    return top++;

}

int ok(char c) //大小写转换

{

    if(c>='a'&&c<='z')

    {

        return c-'a';

    }

    else

    {

        return c-'A';

    }

}

void Build(int Root,int index)

{

    int len = strlen(s[index]);

    for(int i = 0; i < len ;i++)

    {

        int ans = ok(s[index][i]);

        if(Tr[Root].next[ans]==-1)

        {

            Tr[Root].next[ans] = NewNode();

        }

        Root = Tr[Root].next[ans];

    }

    Tr[Root].mark = index;

}

int DFS(int pos)

{

    if(pos==-1)

    {

        return true;

    }

    if(vis[pos]!=-1)//判断是否之前是否遍历到

    {

        return vis[pos];

    }

    int Root = 0;

    for(int i  = pos;i>=0;i--)

    {

        int ans = ok(str[i]);

        if(Tr[Root].next[ans]==-1)

        {

            break;

        }

        Root = Tr[Root].next[ans];

        if(Tr[Root].mark!=-1&&DFS(i-1))

        {

            if( flag ) printf(" ");

            else flag = true;

            printf("%s", s[Tr[Root].mark]);

            return vis[pos] = 1;

        }

    }

    return vis[pos] = 0;

}

int main()

{

    scanf("%d",&n);

    scanf("%s",str);

    scanf("%d",&m);

    top = 0;flag = false;

    int Root = NewNode();

    for(int i=0;i<m;i++)

    {

        scanf("%s",s[i]);

        Build(Root,i);//建立字典树

    }

    for(int i = 0 ; i<MaxM;i++)

    {

        vis[i] = -1;

    }

    DFS(n-1);//记忆化搜索

    return 0;

}