Description

The archaeologists are going to decipher a very mysterious ``language". Now, they know many language patterns; each pattern can be treated as a string on English letters (only lower case). As a sub string, these patterns may appear more than one times in a large text string (also only lower case English letters).

What matters most is that which patterns are the dominating patterns. Dominating pattern is the pattern whose appearing times is not less than other patterns.

It is your job to find the dominating pattern(s) and their appearing times.

Input

Output

Sample Input

2 aba bab ababababac 6 beta alpha haha delta dede tata dedeltalphahahahototatalpha 0

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;#define rep(i,s,t) for(int i=(s);i<(t);i++)const int INF = 1e9 + 9;const int N = 150 * 70 + 9;const int M = 26;char str[155][77];struct Trie {    int nex[N][M], fail[N], val[N];    int rt, L;    int newnode() {        memset (nex[L], -1, sizeof (nex[L]) );        val[L++] = -1;        return L - 1;    }    void init() {        L = 0;        rt = newnode();    }    int idx (char c) {        return c - ‘a‘;    }    void insert (char s[], int id) {        int len = strlen (s);        int now = rt;        for (int i = 0; i < len; i++) {            if (nex[now][idx (s[i])] == -1)                nex[now][idx (s[i])] = newnode();            now = nex[now][idx (s[i])];        }        val[now] = id;    }    void getFail() {        queue<int>q;        fail[rt] = rt;        for (int i = 0; i < M; i++)            if (nex[rt][i] == -1) nex[rt][i] = rt;            else {                fail[nex[rt][i]] = rt;                q.push (nex[rt][i]);            }        while (!q.empty() ) {            int now = q.front();            q.pop();            for (int i = 0; i < M; i++)                if (nex[now][i] == -1)                    nex[now][i] = nex[fail[now]][i];                else {                    fail[nex[now][i]] = nex[fail[now]][i];                    q.push (nex[now][i]);                }        }    }    int num[155];    void quert (char s[], int n) {        memset (num, 0, sizeof (num) );        int len = strlen (s);        int now = rt;        for (int i = 0; i < len; i++) {            now = nex[now][idx(s[i])];            int tmp = now;            while (tmp != rt) {                if (val[tmp] != -1)                    num[val[tmp]]++;                tmp = fail[tmp];            }        }        int maxn=0;        for(int i=0;i<n;i++)maxn=max(num[i],maxn);        printf("%d\n",maxn);        for (int i = 0; i < n; i++)            if (num[i]==maxn) printf ("%s\n", str[i]);    }};char buf[1000009];Trie ac;int main() {   // freopen("f.txt","r",stdin);    int n;    while (~scanf ("%d", &n)&&n ) {        ac.init();        for (int i = 0; i < n; i++) {            scanf ("%s", str[i]);            ac.insert (str[i], i);        }        ac.getFail();        scanf ("%s", buf);        ac.quert (buf, n);    }    return 0;}