#include <bits/stdc++.h>
using namespace std;
#define mod 998244353
int n,m,cnt;
long long pw[305];
bitset<305> bas[305];
bool add(bitset<305> b)
{
	for (int i=m-1;i>=0;i--)
	{
		if (b[i])
		b^=bas[i];
	}
	for (int i=m-1;i>=0;i--)
	{
		if (b[i])
		{
			bas[i]=b;
			return 1;
		}
	}
	return 0;
}
int main()
{
	scanf("%d%d",&n,&m);
	pw[0]=1;
	for (int i=1;i<=300;i++)
	pw[i]=(2*pw[i-1])%mod;
	for (int i=0;i<n;i++)
	{
		bitset<305> cur;
		for (int j=0;j<m;j++)
		{
			int a;
			scanf("%d",&a);
			cur[j]=a;
		}
		cnt+=add(cur);
	}
	printf("%lld",(pw[n]-pw[n-cnt]+mod)%mod*pw[m-1]%mod);
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>
using namespace std;
#define mod 998244353
int n,m,cnt;
long long pw[305];
bitset<305> bas[305];
bool add(bitset<305> b)
{
	for (int i=m-1;i>=0;i--)
	{
		if (b[i])
		b^=bas[i];
	}
	for (int i=m-1;i>=0;i--)
	{
		if (b[i])
		{
			bas[i]=b;
			return 1;
		}
	}
	return 0;
}
int main()
{
	scanf("%d%d",&n,&m);
	pw[0]=1;
	for (int i=1;i<=300;i++)
	pw[i]=(2*pw[i-1])%mod;
	for (int i=0;i<n;i++)
	{
		bitset<305> cur;
		for (int j=0;j<m;j++)
		{
			int a;
			scanf("%d",&a);
			cur[j]=a;
		}
		cnt+=add(cur);
	}
	printf("%lld",(pw[n]-pw[n-cnt]+mod)%mod*pw[m-1]%mod);
}
```