codeforces 633C. Spy Syndrome 2 hash

题目链接

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 


把每一个小的串hash一下， 然后就可以在大的串里面找了， 具体看代码， 很清楚。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
map <ll, string> mp;
ll dp[10005];
void print(int n) {
    if(!n)
        return ;
    string tmp = mp[dp[n]];
    print(n-tmp.size());
    cout<<tmp<<" ";
}
int main()
{
    int n, m;
    string s, str;
    cin>>n>>s>>m;
    for(int i = 0; i<m; i++) {
        cin>>str;
        int len = str.size();
        ll tmp = 0;
        for(int j = 0; j<len; j++) {
            tmp = tmp*31+tolower(str[j]);
        }
        mp[tmp] = str;
    }
    mem1(dp);
    dp[0] = 0;
    int len = s.size();
    for(int i = 0; i<len; i++) {
        ll tmp = 0;
        for(int j = i; j>=0; j--) {
            tmp = tmp*31+s[j];
            if(dp[j]!=-1 && mp.count(tmp)) {
                dp[i+1] = tmp;
                break;
            }
        }
    }
    int i = 0;
    print(len);
    return 0;
}