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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
	public static class bitset {
		long[] a=new long[8];
		public void set(long v,int id) {
			a[id/50]|=v<<(id%50);
		}
		public int get(int id) {
			return (int)((a[id/50]>>(id%50))&1);
		}
		public void xor(bitset rhs) {
			for (int i=0;i<8;++i) a[i]^=rhs.a[i];
		}
	}
	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		int m=in.nextInt();
		bitset[] bit=new bitset[n];
		for (int i=0;i<n;++i) {
			bit[i]=new bitset();
			for (int j=0;j<m;++j) {
				int x=in.nextInt();
				bit[i].set(x,j);
			}
		}
		long mod=998244353;
		long[] p=new long[n+m];
		p[0]=1;
		for (int i=1;i<n+m;++i) p[i]=(p[i-1]+p[i-1])%mod;
		int rk=0;
		for (int i=0;i<m;++i) {
			boolean flag=false;
			for (int j=i;j<n;++j) if (bit[j].get(i)==1) {
				bitset tmp=bit[j];
				bit[j]=bit[i];
				bit[i]=tmp;
				flag=true;
			}
			if (!flag) continue;
			++rk;
			for (int j=i+1;j<n;++j) if (bit[j].get(i)==1)
				bit[j].xor(bit[i]);
		}
		System.out.println((p[n+m-1]-p[n+m-1-rk]+mod)%mod);
	}
}