Keywords Search 

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.

Wiskey also wants to bring this feature to his image retrieval system.

Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.

To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

Input

First line will contain one integer means how many cases will follow by.

Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)

Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.

The last line is the description, and the length will be not longer than 1000000.

 

Output

Print how many keywords are contained in the description.

 

Sample Input

1 5 she he say shr her yasherhs

 

Sample Output

3

#include 
       
         # include
        
          # include
         
          # include
          
             using namespace std;struct node  {      node *fail;       //失败指针      node *next[26]; //Tire每一个节点的26个子节点（最多26个字母）     int count;        //是否为该单词的最后一个节点     node(){           //构造函数初始化         fail=NULL;         count=0;         memset(next,NULL,sizeof(next));     } }*q[1<<6|5]; char keyword[50+5];     //输入的单词 char str[1<<6|5];    //模式串 void insert(char *str, node *root)         //建立字典树 {       node *p=root;    for(int i=0;str[i]!='\0';i++)    {       int  id=str[i]-'a';        if(p->next[id]==NULL)              p->next[id]=new node();        p=p->next[id];     }     p->count++; }void build_ac_automation(node *root){    queue
           
             Q;     int i;     root->fail=NULL;     Q.push(root);     while(!Q.empty())        {          node *temp=Q.front();           Q.pop();           node *p=NULL;           for(i=0;i<26;i++)          {             if(temp->next[i]!=NULL)       //temp 为父结点             {                 if(temp==root) temp->next[i]->fail=root;                 else                 {                     p=temp->fail;   //  思路的关键点，                     while(p!=NULL)                    {                         if(p->next[i]!=NULL)                         {                             temp->next[i]->fail=p->next[i];                             break;                         }                         p=p->fail;    //p=p->fail也就是p=NULL                     }                     if(p==NULL)    temp->next[i]->fail=root;                 }                 Q.push(temp->next[i]);             }         }     } } int query(node *root) {     int i=0,cnt=0,len=strlen(str);     node *p=root;    for(int i=0;i
            
             next[id]==NULL && p!=root) p=p->fail; p=p->next[id]; p=(p==NULL)?root:p; node *temp=p; while(temp!=root && temp->count!=-1) { cnt+=temp->count; temp->count=-1; //表示该单词已经出现过了。防止反复计数 temp=temp->fail; //temp指向e节点的失败指针所指向的节点继续查找 } } return cnt; } int main() { int n,t; scanf("%d",&t); while(t--) { node *root=new node(); scanf("%d",&n); getchar(); while(n--) { gets(keyword); insert(keyword,root); } build_ac_automation(root); scanf("%s",str); printf("%d\n",query(root)); } return 0;}