#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<bitset>
using namespace std;
typedef long long LL;
const int MOD=998244353;
const int N=305;
int n,m;
bitset<N> s[N];
bitset<N> f[N];
int cnt=0;
void Ins ()
{
	for (int u=1;u<=n;u++)
	{
		bool tf=true;
		for (int i=1;i<=m;i++)
		if (s[u][i])
		{
			if (!f[i][i]) 	{f[i]=s[u];tf=false;break;}
			s[u]^=f[i];
		}
		cnt=cnt+tf;
	}
}
int Pow[N];
int main()
{
	scanf("%d%d",&n,&m);
	Pow[0]=1;for (int u=1;u<=n;u++) Pow[u]=Pow[u-1]*2%MOD;
	for (int u=1;u<=n;u++)
	{
		for (int i=1;i<=m;i++)
		{
			int x;
			scanf("%d",&x);
			s[u][i]=x;
		}
	}
	Ins();
	int ans=0;
	int t=1;	for (int u=1;u<m;u++) t=t*2%MOD;
	int lalal=(Pow[n]-Pow[cnt]+MOD)%MOD;
	//printf("%d %d\n",lalal,Pow[cnt]);
	ans=(LL)lalal*t%MOD;
	printf("%d\n",ans%MOD);
	return 0;
}
### 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<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<bitset>
using namespace std;
typedef long long LL;
const int MOD=998244353;
const int N=305;
int n,m;
bitset<N> s[N];
bitset<N> f[N];
int cnt=0;
void Ins ()
{
	for (int u=1;u<=n;u++)
	{
		bool tf=true;
		for (int i=1;i<=m;i++)
		if (s[u][i])
		{
			if (!f[i][i]) 	{f[i]=s[u];tf=false;break;}
			s[u]^=f[i];
		}
		cnt=cnt+tf;
	}
}
int Pow[N];
int main()
{
	scanf("%d%d",&n,&m);
	Pow[0]=1;for (int u=1;u<=n;u++) Pow[u]=Pow[u-1]*2%MOD;
	for (int u=1;u<=n;u++)
	{
		for (int i=1;i<=m;i++)
		{
			int x;
			scanf("%d",&x);
			s[u][i]=x;
		}
	}
	Ins();
	int ans=0;
	int t=1;	for (int u=1;u<m;u++) t=t*2%MOD;
	int lalal=(Pow[n]-Pow[cnt]+MOD)%MOD;
	//printf("%d %d\n",lalal,Pow[cnt]);
	ans=(LL)lalal*t%MOD;
	printf("%d\n",ans%MOD);
	return 0;
}
```