#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n,m,i,j,k,u,mat[305][305],ans=0;
ll mod=998244353; 
ll pow(ll a,ll b)
{
	ll ans=1;
	while(b)
	{
		if(b&1)ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%lld%lld",&n,&m);
	for(i=1;i<=n;i++)for(j=1;j<=m;j++)scanf("%lld",&mat[i][j]);
	for(i=1;i<=m;i++)
	{
		for(j=ans+1;j<=n;j++)
		{
			if(mat[j][i])
			{
				ans++;
				for(k=j+1;k<=n;k++)if(mat[k][i])for(u=i;u<=m;u++)mat[k][u]=mat[k][u]^mat[j][u];
				for(k=i;k<=m;k++)swap(mat[j][k],mat[ans][k]);
				break;
			}
		}
	}
	printf("%lld\n",(pow(2LL,m+n-1)-pow(2LL,m-ans+n-1)+mod)%mod);
	return 0;
}
### Prompt

Generate 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<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n,m,i,j,k,u,mat[305][305],ans=0;
ll mod=998244353; 
ll pow(ll a,ll b)
{
	ll ans=1;
	while(b)
	{
		if(b&1)ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%lld%lld",&n,&m);
	for(i=1;i<=n;i++)for(j=1;j<=m;j++)scanf("%lld",&mat[i][j]);
	for(i=1;i<=m;i++)
	{
		for(j=ans+1;j<=n;j++)
		{
			if(mat[j][i])
			{
				ans++;
				for(k=j+1;k<=n;k++)if(mat[k][i])for(u=i;u<=m;u++)mat[k][u]=mat[k][u]^mat[j][u];
				for(k=i;k<=m;k++)swap(mat[j][k],mat[ans][k]);
				break;
			}
		}
	}
	printf("%lld\n",(pow(2LL,m+n-1)-pow(2LL,m-ans+n-1)+mod)%mod);
	return 0;
}
```