#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll Mod=998244353;
const int maxn=310;
int n,m;
bitset<maxn> p[maxn],b;
ll pw(ll x,ll y)
{
	ll res=1;
	while (y)
	{
		if (y&1)
		{
			res=res*x%Mod;
		}
		x=x*x%Mod;
		y>>=1;
	}
	return res;
}
int main()
{
	scanf("%d%d",&n,&m);
	int r=0;
	for (int i=1;i<=n;i++)
	{
		b.reset();
		for (int j=1;j<=m;j++)
		{
			int x;
			scanf("%d",&x);
			if (x)
			{
				b.set(j);
			}
		}
		for (int j=m;j;j--)
		{
			if (!b[j])
			{
				continue;
			}
			if (!p[j].any())
			{
				p[j]=b;
				break;
			}
			b^=p[j];
		}
		if (!b.any())
		{
			r++;
		}
	}
	cout<<(pw(2,n)-pw(2,r)+Mod)%Mod*pw(2,m-1)%Mod<<endl;
	return 0;
}
### Prompt

Please create a solution in Cpp 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;
#define ll long long
const ll Mod=998244353;
const int maxn=310;
int n,m;
bitset<maxn> p[maxn],b;
ll pw(ll x,ll y)
{
	ll res=1;
	while (y)
	{
		if (y&1)
		{
			res=res*x%Mod;
		}
		x=x*x%Mod;
		y>>=1;
	}
	return res;
}
int main()
{
	scanf("%d%d",&n,&m);
	int r=0;
	for (int i=1;i<=n;i++)
	{
		b.reset();
		for (int j=1;j<=m;j++)
		{
			int x;
			scanf("%d",&x);
			if (x)
			{
				b.set(j);
			}
		}
		for (int j=m;j;j--)
		{
			if (!b[j])
			{
				continue;
			}
			if (!p[j].any())
			{
				p[j]=b;
				break;
			}
			b^=p[j];
		}
		if (!b.any())
		{
			r++;
		}
	}
	cout<<(pw(2,n)-pw(2,r)+Mod)%Mod*pw(2,m-1)%Mod<<endl;
	return 0;
}
```