Problem:
Score : 300 points

Problem Statement

We have N switches with "on" and "off" state, and M bulbs. The switches are numbered 1 to N, and the bulbs are numbered 1 to M.

Bulb i is connected to k_i switches: Switch s_{i1}, s_{i2}, ..., and s_{ik_i}. It is lighted when the number of switches that are "on" among these switches is congruent to p_i modulo 2.

How many combinations of "on" and "off" states of the switches light all the bulbs?

Constraints

1 \leq N, M \leq 10

1 \leq k_i \leq N

1 \leq s_{ij} \leq N

s_{ia} \neq s_{ib} (a \neq b)

p_i is 0 or 1.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
k_1 s_{11} s_{12} ... s_{1k_1}
:
k_M s_{M1} s_{M2} ... s_{Mk_M}
p_1 p_2 ... p_M

Output

Print the number of combinations of "on" and "off" states of the switches that light all the bulbs.

Sample Input 1

2 2
2 1 2
1 2
0 1

Sample Output 1

1

Bulb 1 is lighted when there is an even number of switches that are "on" among the following: Switch 1 and 2.

Bulb 2 is lighted when there is an odd number of switches that are "on" among the following: Switch 2.

There are four possible combinations of states of (Switch 1, Switch 2): (on, on), (on, off), (off, on) and (off, off). Among them, only (on, on) lights all the bulbs, so we should print 1.

Sample Input 2

2 3
2 1 2
1 1
1 2
0 0 1

Sample Output 2

0

Bulb 1 is lighted when there is an even number of switches that are "on" among the following: Switch 1 and 2.

Bulb 2 is lighted when there is an even number of switches that are "on" among the following: Switch 1.

Bulb 3 is lighted when there is an odd number of switches that are "on" among the following: Switch 2.

Switch 1 has to be "off" to light Bulb 2 and Switch 2 has to be "on" to light Bulb 3, but then Bulb 1 will not be lighted. Thus, there are no combinations of states of the switches that light all the bulbs, so we should print 0.

Sample Input 3

5 2
3 1 2 5
2 2 3
1 0

Sample Output 3

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

//////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////// ここからコピペ ////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////

/////////////////////////
// 2進数体の行列のランク //
/////////////////////////

// M×N型行列
template<int M, int N>
class GaussianElimination{
private:
	using Matrix = std::array<std::bitset<N>, M>;

	std::array<int, M> accessIndex;

	bool keyExist(const int key_r, const int key_c)
	{
		for (int r_i{key_r}; r_i < M; r_i++)
			if (matrix_[accessIndex[r_i]][key_c])
			{
				std::swap(accessIndex[key_r], accessIndex[r_i]);
				return true;
			}
		return false;
	}

public:
	Matrix matrix_;
	int rank_;
	GaussianElimination(const Matrix& matrix)
		: matrix_(matrix)
	{
		for (int i{}; i < M; i++) accessIndex[i] = i;
		int key_r{};
		for (int key_c{}; key_r < N && key_c < M; key_r++, key_c++)
		{
			while (key_c < N && !keyExist(key_r, key_c))
				key_c++;
			if (key_c == N) break;
			for (int r_i{}; r_i < M; r_i++)
				if (r_i != key_r && matrix_[accessIndex[r_i]][key_c])
					matrix_[accessIndex[r_i]] ^= matrix_[accessIndex[key_r]];
		}
		rank_ = key_r;
	}
};

//////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////// ここまでコピペ ////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////

int main()
{
	int N, M;
	scanf("%d%d", &N, &M);
	std::array<std::bitset<11>, 10> matrix;
	for (int i{}; i < M; i++)
	{
		int k;
		scanf("%d", &k);
		for (int j{}; j < k; j++)
		{
			int s;
			scanf("%d", &s);
			s--;
			matrix[i][s] = true;
		}
	}
	for (int i{}; i < M; i++)
	{
		int p;
		scanf("%d", &p);
		matrix[i][N] = p == 1;
	}
	GaussianElimination<10, 11> gauss(matrix);
	bool incons{true};
	if (gauss.rank_ > 0)
		for (int i{}; i < N; i++)
			if (gauss.matrix_[gauss.rank_ - 1][i])
				incons = false;
	incons &= gauss.matrix_[gauss.rank_ - 1][N];
	if (incons)
		puts("0");
	else
	{
		int ans{1};
		for (int i{}; i < N - gauss.rank_; i++) ans *= 2;
		printf("%d\n", ans);
	}
	return 0;
}