Welcome to ICPCCamp 2017

ICPCCamp teams are often selected by a mysterious $(X, Y)$-rule described in a blog (?).

There are $(n + 1)$ selection contests held to choose ICPCCamp team among $m$ teams labeled $1, 2, \dots, m$. The number of teams attending the $i$-th contest is $k_i$. The last contest, the $(n + 1)$-th, is EasyCamp-Final, which is very important, so $k_{n + 1} = m$ always holds. The scoreboard of the $i$-th contest is $r_{i, 1}, r_{i, 2}, \dots, r_{i, k_i}$, which means team $r_{i, j}$ has rank $j$ in that contest.

The $(X, Y)$-rule works as follows. First, two non-negative integers $X$ and $Y$ and a permutation $P = \{p_1, p_2, \dots, p_n\}$ of $\{1, 2, \dots, n\}$ are chosen. Then the first $X+Y$ distinct teams in the list $\{r_{n + 1, 1}, \dots, r_{n + 1, Y}, r_{p_1, 1}, r_{p_2, 1}, \dots, r_{p_n, 1}, r_{p_1, 2}, r_{p_2, 2}, \dots, r_{p_n, 2}, \dots\}$ are selected as the ICPCCamp team. In other words, the list starts with the first $Y$ EasyCamp-Final teams, then the top teams of the first $n$ contests in the order given by $P$, then the second teams of the first $n$ contests in the same order, and so on.

Bobo would like to know the number of possible sets of ICPCCamp team modulo $10^9+7$, if he can choose $X$, $Y$, and $P$ arbitrarily.

Wish you enjoy yourself in the upcoming World Finals!

The input contains zero or more test cases and ends at end-of-file. Each test case is given as follows.

The first line contains two integers $n$ and $m$ ($0 \leq n \leq 2 \cdot 10^5$, $1 \leq m \leq 2 \cdot 10^5$).

The $i$-th of the following $n$ lines contains an integer $k_i$ followed by $k_i$ integers $r_{i, 1}, r_{i, 2}, \dots, r_{i, k_i}$ ($1 \leq k_i \leq m$).

The last line contains $m$ integers $r_{n + 1, 1}, r_{n + 1, 2}, \dots, r_{n + 1, m}$ ($1 \leq r_{i, j} \leq m$). For each $i$, the numbers $r_{i, 1}, \dots, r_{i, k_i}$ are distinct.

The sum of $k_i$ and the sum of $m$ over all test cases do not exceed $2 \cdot 10^5$.

For each test case, output one integer, the number of possible sets modulo $10^9+7$.