CF1466F.Euclid's nightmare

You may know that Euclid was a mathematician. Well, as it turns out, Morpheus knew it too. So when he wanted to play a mean trick on Euclid, he sent him an appropriate nightmare.

In his bad dream Euclid has a set $S$ of $n$ $m$-dimensional vectors over the $\mathbb{Z}_2$ field and can perform vector addition on them. In other words he has vectors with $m$ coordinates, each one equal either $0$ or $1$. Vector addition is defined as follows: let $u+v = w$, then $w_i = (u_i + v_i) \bmod 2$.

Euclid can sum any subset of $S$ and archive another $m$-dimensional vector over $\mathbb{Z}_2$. In particular, he can sum together an empty subset; in such a case, the resulting vector has all coordinates equal $0$.

Let $T$ be the set of all the vectors that can be written as a sum of some vectors from $S$. Now Euclid wonders the size of $T$ and whether he can use only a subset $S'$ of $S$ to obtain all the vectors from $T$. As it is usually the case in such scenarios, he will not wake up until he figures this out. So far, things are looking rather grim for the philosopher. But there is hope, as he noticed that all vectors in $S$ have at most $2$ coordinates equal $1$.

Help Euclid and calculate $|T|$, the number of $m$-dimensional vectors over $\mathbb{Z}_2$ that can be written as a sum of some vectors from $S$. As it can be quite large, calculate it modulo $10^9+7$. You should also find $S'$, the smallest such subset of $S$, that all vectors in $T$ can be written as a sum of vectors from $S'$. In case there are multiple such sets with a minimal number of elements, output the lexicographically smallest one with respect to the order in which their elements are given in the input.

Consider sets $A$ and $B$ such that $|A| = |B|$. Let $a_1, a_2, \dots a_{|A|}$ and $b_1, b_2, \dots b_{|B|}$ be increasing arrays of indices elements of $A$ and $B$ correspondingly. $A$ is lexicographically smaller than $B$ iff there exists such $i$ that $a_j = b_j$ for all $j \lt i$ and $a_i \lt b_i$.

你可能知道欧几里得是一位数学家。事实上，墨菲斯也知道这一点。因此，当他想对欧几里得施加一个恶作剧时，便给他安排了一场恰如其分的噩梦。

在噩梦中，欧几里得拥有一个由 $n$ 个 $m$ 维向量构成的集合 $S$，这些向量定义在 $\mathbb{Z}_2$ 域上，并可在其上进行向量加法运算。换言之，这些向量具有 $m$ 个坐标，每个坐标值非 $0$ 即 $1$。向量加法定义如下：设 $u+v = w$，则 $w_i = (u_i + v_i) \bmod 2$。

欧几里得可以对 $S$ 的任意子集求和，从而得到另一个 $m$ 维 $\mathbb{Z}_2$ 向量。特别地，他也可以对空子集求和；此时所得向量的所有坐标均为 $0$。

令 $T$ 表示所有可表示为 $S$ 中某些向量之和的向量所构成的集合。现在欧几里得想知道 $T$ 的大小 $|T|$，以及是否能仅用 $S$ 的某个子集 $S'$ 就生成 $T$ 中全部向量。正如这类情形中通常发生的那样，他只有彻底弄清这个问题后才能醒来。目前，这位哲学家的处境看起来相当不妙。但尚存一线希望——他注意到 $S$ 中所有向量至多有 $2$ 个坐标等于 $1$。

请帮助欧几里得计算 $|T|$，即所有可表示为 $S$ 中某些向量之和的 $m$ 维 $\mathbb{Z}_2$ 向量的个数。由于该值可能非常大，请对 $10^9+7$ 取模输出。你还需找出 $S'$，即满足“利用 $S'$ 中向量的和即可生成 $T$ 中所有向量”的最小（元素个数最少）子集 $S' \subseteq S$。若存在多个满足最小元素个数的子集，则输出其中按输入顺序下标字典序最小的那个。

考虑两个集合 $A$ 和 $B$，且 $|A| = |B|$。设 $a_1, a_2, \dots, a_{|A|}$ 和 $b_1, b_2, \dots, b_{|B|}$ 分别为 $A$ 和 $B$ 中元素在输入中下标的升序排列数组。称 $A$ 字典序小于 $B$，当且仅当存在某个 $i$，使得对所有 $j < i$ 都有 $a_j = b_j$，且 $a_i < b_i$。

In the first line of input, there are two integers $n$, $m$ ($1 \leq n, m \leq 5 \cdot 10^5$) denoting the number of vectors in $S$ and the number of dimensions.

Next $n$ lines contain the description of the vectors in $S$. In each of them there is an integer $k$ ($1 \leq k \leq 2$) and then follow $k$ distinct integers $x_1, \dots x_k$ ($1 \leq x_i \leq m$). This encodes an $m$-dimensional vector having $1$s on coordinates $x_1, \dots x_k$ and $0$s on the rest of them.

Among the $n$ vectors, no two are the same.

输入的第一行包含两个整数 $n$ 和 $m$（$1 \leq n, m \leq 5 \cdot 10^5$），分别表示集合 $S$ 中向量的个数以及向量的维数。

接下来的 $n$ 行描述集合 $S$ 中的向量。每行首先是一个整数 $k$（$1 \leq k \leq 2$），随后是 $k$ 个互不相同的整数 $x_1, \dots, x_k$（$1 \leq x_i \leq m$）。这表示一个 $m$ 维向量，其在坐标 $x_1, \dots, x_k$ 处取值为 $1$，其余坐标处取值为 $0$。

在全部 $n$ 个向量中，任意两个向量均不相同。

In the first line, output two integers: remainder modulo $10^9+7$ of $|T|$ and $|S'|$. In the second line, output $|S'|$ numbers, indices of the elements of $S'$ in ascending order. The elements of $S$ are numbered from $1$ in the order they are given in the input.

第一行输出两个整数：$|T|$ 对 $10^9+7$ 取模的余数，以及 $|S'|$。
第二行输出 $|S'|$ 个数字，表示 $S'$ 中各元素在 $S$ 中的下标（按升序排列）。$S$ 中的元素从 $1$ 开始编号，编号顺序与输入中给出的顺序一致。

In the first example we are given three vectors:

• $10$
• $01$
• $11$

It turns out that we can represent all vectors from our $2$-dimensional space using these vectors:

• $00$ is a sum of the empty subset of above vectors;
• $01 = 11 + 10$, is a sum of the first and third vector;
• $10 = 10$, is just the first vector;
• $11 = 10 + 01$, is a sum of the first and the second vector.

Hence, $T = {00, 01, 10, 11}$. We can choose any two of the three vectors from $S$ and still be able to obtain all the vectors in $T$. In such a case, we choose the two vectors which appear first in the input. Since we cannot obtain all vectors in $T$ using only a single vector from $S$, $|S'| = 2$ and $S' = {10, 01}$ (indices $1$ and $2$), as set ${1, 2 }$ is lexicographically the smallest. We can represent all vectors from $T$, using only vectors from $S'$, as shown below:

• $00$ is a sum of the empty subset;
• $01 = 01$ is just the second vector;
• $10 = 10$ is just the first vector;
• $11 = 10 + 01$ is a sum of the first and the second vector.

在第一个例子中，我们给出了三个向量：

• $10$
• $01$
• $11$

事实上，我们可以利用这三个向量表示二维空间中的所有向量：

• $00$ 是上述向量的空子集的和；
• $01 = 11 + 10$，是第一个与第三个向量的和；
• $10 = 10$，即第一个向量本身；
• $11 = 10 + 01$，是第一个与第二个向量的和。

因此，$T = \{00, 01, 10, 11\}$。我们可以从 $S$ 中任选其中两个向量，仍能生成 $T$ 中的所有向量。此时，我们选择输入中出现顺序最靠前的两个向量。由于仅用 $S$ 中的一个向量无法生成 $T$ 中的所有向量，故 $|S'| = 2$，且 $S' = \{10, 01\}$（对应下标 $1$ 和 $2$），因为集合 $\{1, 2\}$ 是字典序最小的。我们仅使用 $S'$ 中的向量即可表示 $T$ 中的所有向量，如下所示：

• $00$ 是空子集的和；
• $01 = 01$ 即第二个向量本身；
• $10 = 10$ 即第一个向量本身；
• $11 = 10 + 01$ 是第一个与第二个向量的和。