#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ri register int
const int N=500005,M=998244353;
int n,m,i,j,ans,vis[305];
bitset<305> p[305],a[305];
long long qpow(long long a,long long b)
{
	long long s=a,ans=1;
	while(b)
	{
		if(b&1)
			ans=ans*s%M;
		s=s*s%M;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++)
		for(j=1;j<=m;j++)
		{
			int t;
			scanf("%d",&t);
			a[i][j-1]=t;
		}
	for(i=1;i<=n;i++)
	{
		for(j=m-1;j>=0;--j)
			if(a[i][j])
				if(vis[j]==0)
				{
					p[j]=a[i];
					vis[j]=1;
					break;
				}
				else
					a[i]^=p[j];
		if(j<0)
			ans++;
	}
	cout<<((qpow(2,n)-qpow(2,ans))*qpow(2,m-1)%M+M)%M;
}

### Prompt

Please formulate 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ri register int
const int N=500005,M=998244353;
int n,m,i,j,ans,vis[305];
bitset<305> p[305],a[305];
long long qpow(long long a,long long b)
{
	long long s=a,ans=1;
	while(b)
	{
		if(b&1)
			ans=ans*s%M;
		s=s*s%M;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++)
		for(j=1;j<=m;j++)
		{
			int t;
			scanf("%d",&t);
			a[i][j-1]=t;
		}
	for(i=1;i<=n;i++)
	{
		for(j=m-1;j>=0;--j)
			if(a[i][j])
				if(vis[j]==0)
				{
					p[j]=a[i];
					vis[j]=1;
					break;
				}
				else
					a[i]^=p[j];
		if(j<0)
			ans++;
	}
	cout<<((qpow(2,n)-qpow(2,ans))*qpow(2,m-1)%M+M)%M;
}

```