#include<bits/stdc++.h>
using namespace std;
const int mod=998244353;
int n,m,ans,A[305][305],pw[605];
int guass(int m,int n)
{
	int i=0,j=0,r;
	while(i<m&&j<n)
	{
		r=i;
		for(int k=i;k<m;k++)if(A[k][j]){r=k;break;}
		if(A[r][j])
		{
			if(r!=i)for(int k=0;k<=n;k++)swap(A[r][k],A[i][k]);
			for(int u=i+1;u<m;u++)
			if(A[u][j])for(int k=i;k<=n;k++)A[u][k]^=A[i][k];
			i++;
		}
		j++;
	}
	return i;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
	for(int j=0;j<m;j++)
	scanf("%d",&A[j][i]);
	pw[0]=1;
	for(int i=1;i<=n+m;i++)pw[i]=2ll*pw[i-1]%mod;
	ans=(pw[n+m-1]-pw[n+m-1-guass(m,n)]+mod)%mod; 
	printf("%d\n",ans);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<bits/stdc++.h>
using namespace std;
const int mod=998244353;
int n,m,ans,A[305][305],pw[605];
int guass(int m,int n)
{
	int i=0,j=0,r;
	while(i<m&&j<n)
	{
		r=i;
		for(int k=i;k<m;k++)if(A[k][j]){r=k;break;}
		if(A[r][j])
		{
			if(r!=i)for(int k=0;k<=n;k++)swap(A[r][k],A[i][k]);
			for(int u=i+1;u<m;u++)
			if(A[u][j])for(int k=i;k<=n;k++)A[u][k]^=A[i][k];
			i++;
		}
		j++;
	}
	return i;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
	for(int j=0;j<m;j++)
	scanf("%d",&A[j][i]);
	pw[0]=1;
	for(int i=1;i<=n+m;i++)pw[i]=2ll*pw[i-1]%mod;
	ans=(pw[n+m-1]-pw[n+m-1-guass(m,n)]+mod)%mod; 
	printf("%d\n",ans);
}
```