/**

* ZOJ 1140 Courses

*

* 00:00.73 964k

*/

#include <iostream>

#include <string>

#include <cstdio>

using namespace std;

//二分图完备匹配,hungary算法,邻接阵形式,复杂度O(m*m*n)

//返回是否是完备匹配,传入二分图大小m,n和邻接阵mat,非零元素表示有边

//如存在,match1,match2返回一个完备匹配,未匹配顶点match值为-1,否则无意义

#define MAXN 300

#define _clr(x) memset(x,0xff,sizeof(int)*MAXN)

int hungary(int m,int n,int mat[][MAXN],int* match1,int* match2){

int s[MAXN],t[MAXN],p,q,ret=0,i,j,k;

if (m>n) return 0;

for (_clr(match1),_clr(match2),i=0;i<m&&ret==i;ret+=(match1[i++]>=0))

for (_clr(t),s[p=q=0]=i;p<=q&&match1[i]<0;p++)

for (k=s[p],j=0;j<n&&match1[i]<0;j++)

if (mat[k][j]&&t[j]<0){

s[++q]=match2[j],t[j]=k;

if (s[q]<0)

for (p=j;p>=0;j=p)

match2[j]=k=t[j],p=match1[k],match1[k]=j;

}

return ret==m;

}

int match1[MAXN], match2[MAXN];

int m, n;

int mat[100][300];

int Run() {

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

memset(mat, 0, sizeof(mat));

for (int i = 0; i < m; ++i) {

int num, tmp;

scanf("%d", &num);

for (int j = 0; j < num; ++j) {

scanf("%d", &tmp);

--tmp;

mat[i][tmp] = 1;

}

}

if (hungary(m, n, mat, match1, match2))

printf("YES\n");

else

printf("NO\n");

return 0;

}

int main() {

int kase;

for (scanf("%d", &kase); kase; --kase) {

Run();

}

return 0;

}