#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<bitset>

typedef long long LL;

const int N=305;
const int MOD=998244353;

int n,m,tot;
bool vis[N];
std::bitset<N> bit[N],bas[N];

int ksm(int x,int y)
{
	int ans=1;
	while (y)
	{
		if (y&1) ans=(LL)ans*x%MOD;
		x=(LL)x*x%MOD;y>>=1;
	}
	return ans;
}

void gauss()
{
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			if (bit[i][j])
			{
				if (vis[j]) bit[i]^=bas[j];
				else
				{
					bas[j]=bit[i];
					tot++;
					vis[j]=1;
					break;
				}
			}
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
		{
			int x;scanf("%d",&x);
			bit[i][j]=x;
		}
	gauss();
	int ans=(LL)ksm(2,n-1)*(ksm(2,m)-ksm(2,m-tot))%MOD;
	ans+=ans<0?MOD:0;
	printf("%d\n",ans);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<bitset>

typedef long long LL;

const int N=305;
const int MOD=998244353;

int n,m,tot;
bool vis[N];
std::bitset<N> bit[N],bas[N];

int ksm(int x,int y)
{
	int ans=1;
	while (y)
	{
		if (y&1) ans=(LL)ans*x%MOD;
		x=(LL)x*x%MOD;y>>=1;
	}
	return ans;
}

void gauss()
{
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			if (bit[i][j])
			{
				if (vis[j]) bit[i]^=bas[j];
				else
				{
					bas[j]=bit[i];
					tot++;
					vis[j]=1;
					break;
				}
			}
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
		{
			int x;scanf("%d",&x);
			bit[i][j]=x;
		}
	gauss();
	int ans=(LL)ksm(2,n-1)*(ksm(2,m)-ksm(2,m-tot))%MOD;
	ans+=ans<0?MOD:0;
	printf("%d\n",ans);
	return 0;
}
```