# Spy Syndrome 2

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 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 w i (|w i| ≤ 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

30ariksihsidlihcdnaehsetahgnisol10KirahatesishelosingdeathchildishLandNote

output

Kira is childish and he hates losing

input

12iherehtolleh5HIHothereHeLLohello

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.

#include <bits/stdc++.h>

using namespace std;

const int N = 1000010;

#define end END

int a[N];
char str[N];
int end[N];
int pr[N], pw[N];
char word[N];
string name[N];

int main() {
int len;
scanf("%d", &len);
scanf("%s", str);
int cnt;
scanf("%d", &cnt);
int nn = 1;
end = -1;
memset(a, 0, sizeof(a));
for (int i = 0; i < cnt; i++) {
scanf("%s", word);
name[i] = "";
for (int j = 0; word[j]; j++) {
name[i] += word[j];
}
int t = 1;
for (int j = 0; word[j]; j++) {
if (word[j] >= 'A' && word[j] <= 'Z') {
word[j] += 32;
}
int c = word[j] - 'a';
if (a[t] == 0) {
nn++;
end[nn] = -1;
a[t] = nn;
}
t = a[t];
}
end[t] = i;
}
for (int i = 0; i <= len; i++) {
pr[i] = -1;
pw[i] = -1;
}
pr[len] = 0;
pw[len] = 0;
for (int i = len; i > 0; i--) {
if (pr[i] == -1) {
continue;
}
int t = 1;
for (int j = i - 1; j >= 0; j--) {
int c = str[j] - 'a';
if (a[t] == 0) {
break;
}
t = a[t];
if (end[t] != -1) {
pr[j] = i;
pw[j] = end[t];
}
}
}
int id = 0;
while (id < len) {
if (id > 0) {
putchar(' ');
}
printf("%s", name[pw[id]].c_str());
id = pr[id];
}
printf("\n");
return 0;
}