#include <stdio.h>
#include <algorithm>

using namespace std;

const long long maxn = 310;
const long long mod = 998244353;
long long n , m;
long long a[maxn][maxn];
long long s[maxn][maxn];
long long r;
long long pow ( long long x ) {
	long long s = 1 , f = 2;
	while ( x ) {
		if ( x % 2 ) s = ( s * f ) % mod;
		f = ( f * f ) % mod; x = x / 2;
	}
	return s;
}
void work () {
	long long i , j , k;
	scanf ( "%lld%lld" , &n , &m );
	for ( i = 1 ; i <= n ; i++ )
		for ( j = 1 ; j <= m ; j++ )
			scanf ( "%lld" , &a[i][j] );
	for ( i = 1 ; i <= n ; i++ ) {
		for ( j = 1 ; j <= m ; j++ ) {
			if ( s[j][j] == 0 && a[i][j] == 1 ) {
				for ( k = 1 ; k <= m ; k++ ) {
					s[j][k] = a[i][k];
				}
				r++;
				break;
			}
			if ( a[i][j] == 1 )
				for ( k = 1 ; k <= m ; k++ ) {
					a[i][k] ^= s[j][k];
				}
		}
	}
	printf ( "%lld\n" , ( pow ( n + m - 1 ) - pow ( n + m - r - 1 ) + mod ) % mod );
}
int main () {
	work ();
	return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

Examples

Input

2 2
0 1
1 0


Output

6


Input

2 3
0 0 0
0 1 0


Output

8

### Response

```cpp
#include <stdio.h>
#include <algorithm>

using namespace std;

const long long maxn = 310;
const long long mod = 998244353;
long long n , m;
long long a[maxn][maxn];
long long s[maxn][maxn];
long long r;
long long pow ( long long x ) {
	long long s = 1 , f = 2;
	while ( x ) {
		if ( x % 2 ) s = ( s * f ) % mod;
		f = ( f * f ) % mod; x = x / 2;
	}
	return s;
}
void work () {
	long long i , j , k;
	scanf ( "%lld%lld" , &n , &m );
	for ( i = 1 ; i <= n ; i++ )
		for ( j = 1 ; j <= m ; j++ )
			scanf ( "%lld" , &a[i][j] );
	for ( i = 1 ; i <= n ; i++ ) {
		for ( j = 1 ; j <= m ; j++ ) {
			if ( s[j][j] == 0 && a[i][j] == 1 ) {
				for ( k = 1 ; k <= m ; k++ ) {
					s[j][k] = a[i][k];
				}
				r++;
				break;
			}
			if ( a[i][j] == 1 )
				for ( k = 1 ; k <= m ; k++ ) {
					a[i][k] ^= s[j][k];
				}
		}
	}
	printf ( "%lld\n" , ( pow ( n + m - 1 ) - pow ( n + m - r - 1 ) + mod ) % mod );
}
int main () {
	work ();
	return 0;
}

```