/*
Keywords Search
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 47775 Accepted Submission(s): 15220
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.
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.
*/
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int size,a[500001][27],T,shu[500001],dui[500001],fail[500001],sum;
bool f[500001];
char ch[100],s[1000008];
void jia()
{
int l=strlen(ch),now=1;
for(int i=0;i<l;i++)
if(a[now][ch[i]-'a'+1])
now=a[now][ch[i]-'a'+1];
else
{
size++;
now=a[now][ch[i]-'a'+1]=size;
}
shu[now]++;
return;
}
void shi()
{
dui[1]=1;
int h=0,t=1;
for(;h<t;)
{
h++;
int now=dui[h];
for(int i=1;i<=26;i++)
if(a[now][i])
{
int k=fail[now];
for(;!a[k][i];k=fail[k]);
fail[a[now][i]]=a[k][i];
t++;
dui[t]=a[now][i];
}
}
}
void jin()
{
int now=1,l=strlen(s);
for(int i=0;i<l;i++)
{
f[now]=1;
int k=s[i]-'a'+1;
for(;!a[now][k];now=fail[now]);
now=a[now][k];
if(!f[now])
for(int j=now;j;j=fail[j])
{
sum+=shu[j];
shu[j]=0;
}
}
printf("%d
",sum);
}
int main()
{
for(int i=0;i<=26;i++)
a[0][i]=1;
scanf("%d",&T);
for(;T;T--)
{
size=1;
sum=0;
int n;
scanf("%d",&n);
for(;n;n--)
{
scanf("%s",ch);
jia();
}
shi();
scanf("%s",s);
jin();
for(int i=1;i<=size;i++)
{
f[i]=fail[i]=shu[i]=0;
for(int j=1;j<=26;j++)
a[i][j]=0;
}
}
return 0;
}