#include<iostream>
#include<cstdio>

using namespace std;

const int mo=998244353;

int n,m,a[301][301],r=0,b[301][301];
bool f[301];

int mi(int o,int p)
{
	int u=1;
	while(p>0)
	{
		if(p&1) u=1ll*u*o%mo;
		o=1ll*o*o%mo;
		p>>=1;
	}
	return u;
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
	for(int i=1;i<=n;i++)
		for(int j=m;j>=1;j--)
			if(a[i][j])
			{
				if(!f[j]) 
				{
					for(int k=1;k<=m;k++) b[j][k]=a[i][k];
					f[j]=1;r++;
					break;
				}else
				{
					for(int k=1;k<=m;k++) a[i][k]^=b[j][k];
				}
			}
	cout<<((mi(2,n+m-1)-mi(2,n+m-1-r)+mo)%mo)<<endl;
	return 0;
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

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<iostream>
#include<cstdio>

using namespace std;

const int mo=998244353;

int n,m,a[301][301],r=0,b[301][301];
bool f[301];

int mi(int o,int p)
{
	int u=1;
	while(p>0)
	{
		if(p&1) u=1ll*u*o%mo;
		o=1ll*o*o%mo;
		p>>=1;
	}
	return u;
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
	for(int i=1;i<=n;i++)
		for(int j=m;j>=1;j--)
			if(a[i][j])
			{
				if(!f[j]) 
				{
					for(int k=1;k<=m;k++) b[j][k]=a[i][k];
					f[j]=1;r++;
					break;
				}else
				{
					for(int k=1;k<=m;k++) a[i][k]^=b[j][k];
				}
			}
	cout<<((mi(2,n+m-1)-mi(2,n+m-1-r)+mo)%mo)<<endl;
	return 0;
}
```