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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define int long long
#define p 998244353

int MOD(int a){
	a %= p;
	return a >= 0 ? a : a + p;
}

int power(int a, int N){
	if(N == 0){
		return 1;
	}
	else if(N % 2 == 0){
		return power(MOD(a * a), N / 2);
	}
	else{
		return MOD(a * power(a, N - 1));
	}
}

int rank(int N, int M, int **a){
	int i, j, k, l;
	int **sub = (int **)malloc(sizeof(int *) * N);
	for(i = 0; i < N; i++){
		sub[i] = (int *)malloc(sizeof(int) * M);
		for(j = 0; j < M; j++){
			sub[i][j] = a[i][j];
		}
	}
	for(i = 0, j = 0; i < N && j < M; i++, j++){
//		printf("(i, j) = (%lld, %lld)\n", i, j);
		for(k = i; k < N; k++){
			if(sub[k][j] == 1){
				break;
			}
		}
//		printf("k = %lld\n", k);
		if(k == N){
			i--;
			continue;
		}
		int *tmp = sub[i];
		sub[i] = sub[k];
		sub[k] = tmp;
		for(k = 0; k < N; k++){
			if(i != k && sub[k][j] == 1){
				for(l = j; l < M; l++){
					sub[k][l] ^= sub[i][l];
				}
			}
		}
/*		printf("sub:\n");
		for(k = 0; k < N; k++){
			for(l = 0; l < M; l++){
				printf("%lld ", sub[k][l]);
			}
			printf("\n");
		}
*/	}
/*	printf("sub:\n");
	for(k = 0; k < N; k++){
		for(l = 0; l < M; l++){
			printf("%lld ", sub[k][l]);
		}
		printf("\n");
	}
*/	return i;
}

signed main(){
	int N, M, i, j;
	scanf("%lld%lld", &N, &M);
	int **a = (int **)malloc(sizeof(int *) * N);
	for(i = 0; i < N; i++){
		a[i] = (int *)malloc(sizeof(int) * M);
		for(j = 0; j < M; j++){
			scanf("%lld", &a[i][j]);
		}
	}
	printf("%lld\n", MOD((power(2, N) - power(2, N - rank(N, M, a))) * power(2, M - 1)));
	return 0;
}