Problem:
Score : 800 points

Problem Statement

There is a square grid with N rows and M columns.
Each square contains an integer: 0 or 1. The square at the i-th row from the top and the j-th column from the left contains a_{ij}.

Among the 2^{N+M} possible pairs of a subset A of the rows and a subset B of the columns, find the number of the pairs that satisfy the following condition, modulo 998244353:

The sum of the |A||B| numbers contained in the intersection of the rows belonging to A and the columns belonging to B, is odd.

Constraints

1 \leq N,M \leq 300

0 \leq a_{i,j} \leq 1(1\leq i\leq N,1\leq j\leq M)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
a_{11} ... a_{1M}
:
a_{N1} ... a_{NM}

Output

Print the number of the pairs of a subset of the rows and a subset of the columns that satisfy the condition, modulo 998244353.

Sample Input 1

2 2
0 1
1 0

Sample Output 1

6

For example, if A consists of the first row and B consists of both columns, the sum of the numbers contained in the intersection is 0+1=1.

Sample Input 2

2 3
0 0 0
0 1 0

Sample Output 2

8

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

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <tuple>
template <typename T> constexpr T my_abs(const T &x) { return x < 0 ? -x : x; }
using ll = long long;
constexpr int MOD = 998244353;
inline int quick_pow(int a, int n)
{
	int res = 1;
	while (n)
	{
		if (n & 1)
			res = (ll)res * a % MOD;
		a = (ll)a * a % MOD;
		n >>= 1;
	}
	return res;
}
int fact[305], inv[305], arr[305][305], dp[305];
inline int C(int n, int m) { return (ll)fact[n] * inv[m] % MOD * inv[n - m] % MOD; }
int main()
{
	// freopen("yahoo2019-E.in", "r", stdin);
	int n, m, k = 0, ans = 0;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
			scanf("%d", arr[i] + j);
	}
	fact[0] = inv[0] = 1;
	for (int i = 1; i <= m; i++)
	{
		fact[i] = (ll)fact[i - 1] * i % MOD;
		inv[i] = quick_pow(fact[i], MOD - 2);
	}
	for (int col = 0; col < m; col++)
	{
		int row = k;
		while (row < n && !arr[row][col])
			row++;
		if (row >= n)
			continue;
		std::swap(arr[row], arr[k]);
		for (int i = 0; i < n; i++)
		{
			if (!arr[i][col] || i == k)
				continue;
			for (int j = 0; j < m; j++)
				arr[i][j] ^= arr[row][j];
		}
		k++;
	}
	dp[0] = 1;
	for (int i = 0; i < n; i++)
	{
		int sum = 0;
		for (int j = 0; j < m; j++)
			sum += arr[i][j];
		for (int j = m - sum; j >= 0; j--)
			(dp[j + sum] += dp[j]) %= MOD;
	}
	for (int i = 0; i <= m; i++)
	{
		int coef = 0;
		for (int j = 1; j <= i; j += 2)
			(coef += C(i, j)) %= MOD;
		ans = (ans + (ll)dp[i] * coef % MOD * quick_pow(2, m - i)) % MOD;
	}
	printf("%d\n", ans);
	return 0;
}