#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define pa pair<int,int>
const int Maxn=310;
const int inf=2147483647;
const int mod=998244353;
int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return x*f;
}
int n,m,a[Maxn][Maxn],b[Maxn][Maxn],c[Maxn],cnt=0;
int Pow(int x,int y)
{
	if(!y)return 1;
	int t=Pow(x,y>>1),re=(LL)t*t%mod;
	if(y&1)re=(LL)re*x%mod;
	return re;
}
int main()
{
	n=read(),m=read();
	for(int i=1;i<=n;i++)
	for(int j=1;j<=m;j++)
	a[i][j]=read();
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)c[j]=a[i][j];
		for(int j=1;j<=m;j++)
		if(c[j])
		{
			if(!b[j][j])
			{
				for(int k=j;k<=m;k++)b[j][k]=c[k];
				cnt++;break;
			}
			else
			{
				for(int k=j;k<=m;k++)c[k]^=b[j][k];
			}
		}
	}
	printf("%lld",(LL)(Pow(2,n)-Pow(2,n-cnt)+mod)%mod*Pow(2,m-1)%mod);
}
### Prompt

In CPP, your task is to solve 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;
#define LL long long
#define pa pair<int,int>
const int Maxn=310;
const int inf=2147483647;
const int mod=998244353;
int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return x*f;
}
int n,m,a[Maxn][Maxn],b[Maxn][Maxn],c[Maxn],cnt=0;
int Pow(int x,int y)
{
	if(!y)return 1;
	int t=Pow(x,y>>1),re=(LL)t*t%mod;
	if(y&1)re=(LL)re*x%mod;
	return re;
}
int main()
{
	n=read(),m=read();
	for(int i=1;i<=n;i++)
	for(int j=1;j<=m;j++)
	a[i][j]=read();
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)c[j]=a[i][j];
		for(int j=1;j<=m;j++)
		if(c[j])
		{
			if(!b[j][j])
			{
				for(int k=j;k<=m;k++)b[j][k]=c[k];
				cnt++;break;
			}
			else
			{
				for(int k=j;k<=m;k++)c[k]^=b[j][k];
			}
		}
	}
	printf("%lld",(LL)(Pow(2,n)-Pow(2,n-cnt)+mod)%mod*Pow(2,m-1)%mod);
}
```