//Love and Freedom.
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<bitset>
#define inf 20021225
#define ll long long
#define mdn 998244353
using namespace std;

bitset<310> r[310],a[310]; int n,m;
int gauss()
{
	int w = 0;
	/**for(int i=1,j;i<=min(n,m);i++)
	{
		if(!r[i][i])
		{
			for(j=i+1;j<=n;j++)
				if(r[j][i])	break;
			if(j>n)	continue;
			swap(r[i],r[j]);
		}
		for(j=i+1;j<=n;j++)
			if(r[j][i])	r[j]^=r[i];
	}
	for(int i=1;i<=min(n,m);i++)	if(r[i][i])	w++;*/
	for(int i=1;i<=n;i++)
	{
		for(int j=m;j;j--)
		{
			if(!r[i][j])	continue;
			if(!a[j].any())
			{
				a[j]^=r[i],w++;
				break;
			}
			else	r[i]^=a[j];
		}
	}
	return w;
}
int ksm(int bs,int mi)
{
	int ans = 1;
	while(mi)
	{
		if(mi&1)	ans=(ll)ans*bs%mdn;
		bs=(ll)bs*bs%mdn; mi>>=1;
	}
	return ans;
}
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);
			r[i][j] = x;
		}
	int ans = ksm(2,n+m-1)-ksm(2,n+m-1-gauss());
	printf("%d\n",(ans+mdn)%mdn);
	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
//Love and Freedom.
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<bitset>
#define inf 20021225
#define ll long long
#define mdn 998244353
using namespace std;

bitset<310> r[310],a[310]; int n,m;
int gauss()
{
	int w = 0;
	/**for(int i=1,j;i<=min(n,m);i++)
	{
		if(!r[i][i])
		{
			for(j=i+1;j<=n;j++)
				if(r[j][i])	break;
			if(j>n)	continue;
			swap(r[i],r[j]);
		}
		for(j=i+1;j<=n;j++)
			if(r[j][i])	r[j]^=r[i];
	}
	for(int i=1;i<=min(n,m);i++)	if(r[i][i])	w++;*/
	for(int i=1;i<=n;i++)
	{
		for(int j=m;j;j--)
		{
			if(!r[i][j])	continue;
			if(!a[j].any())
			{
				a[j]^=r[i],w++;
				break;
			}
			else	r[i]^=a[j];
		}
	}
	return w;
}
int ksm(int bs,int mi)
{
	int ans = 1;
	while(mi)
	{
		if(mi&1)	ans=(ll)ans*bs%mdn;
		bs=(ll)bs*bs%mdn; mi>>=1;
	}
	return ans;
}
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);
			r[i][j] = x;
		}
	int ans = ksm(2,n+m-1)-ksm(2,n+m-1-gauss());
	printf("%d\n",(ans+mdn)%mdn);
	return 0;
}
```