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全文第一行输入一个正整数 $T(1\le T\le200)$ ，表示数据组数。

对每组数据，第一行输入两个正整数 $n,m(1\le n,m\le200)$ 。

接下来输入一个 $n\times m$ 的矩阵，第 $i$ 行第 $j$ 列表示第 $i$ 个礼物的第 $j$ 种颜色 $col(1\le col\le200)$ 。

接下来输入一个长度为 $m$ 的序列，表示序列 $b(1\le b_i\le200)$ 。

数据保证同一行之内 $col$ 互不相同（即一个礼物上不会有两种相同的颜色），且信息 $b_i$ 互不相同。

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import java.util.Scanner;
public class Main {
    public static void main(String[] arg) {
        Scanner scanner = new Scanner(System.in);
        // todo
    }
}
 
 
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for cas in range(int(input())):

n, m = map(int, input().split())

c = []

for i in range(n):

c.append(list(map(int, input().split())))

a = [i for i in range(n)]

for tmp in input().split():

b = int(tmp)

nxt = []

for x in a:

for y in c[x]:

if y == b:

nxt.append(x)

break

print(len(nxt), end = ' ')

a = nxt

print('')

#include<bits/stdc++.h>

using namespace std;

int main(){

int t,n,m,i,j,v[205][205],x,s;

scanf("%d",&t);

while(t--){

memset(v,0,sizeof v);

scanf("%d%d",&n,&m);

for(i=1;i<=n;i++){

for(j=1;j<=m;j++){

scanf("%d",&x);

v[i][x]=1;

}

for(i=1;i<=m;i++){

s=0;

scanf("%d",&x);

for(j=1;j<=n;j++){

if(!v[j][0]){

if(v[j][x]) s++;

else v[j][0]=1;

}

printf("%d ",s);

}

printf("\n");

}

return 0;

}

T=int(input())

for _ in range(T):

n,m=map(int,input().split(' '))

arr=[1]*n

brr=[[0]*205 for _ in range(n)]

for i in range(n):

for ci in map(int,input().split(' ')):

brr[i][ci]=1

cnt=0

for cj in map(int,input().split(' ')):

for i in range(n):

if brr[i][cj]==0:

arr[i]=0

cnt+=1

print(sum(arr),end=' ')

print()

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