#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<ctime>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<bitset>

#define N 305
#define M 8000005

#define ls (t<<1)
#define rs ((t<<1)|1)
#define mid ((l+r)>>1)

#define mk make_pair
#define pb push_back
#define fi first
#define se second

#define Mo 998244353

using namespace std;

bitset<305>A[305];

int a[N][N],n,m,i,j;

int power(int x,int y)
{
		int s=1;
		for(;y;y>>=1)
		{
				if(y&1) s=1ll*s*x%Mo;
				x=1ll*x*x%Mo; 
		}
		return s;
}

int Guass()
{
		int i,j,k,cnt=0;
		for (i=1;i<=n;++i)
		{
				for (j=cnt+1;j<=m;++j) if (A[j][i]) break;
				if (j>m) continue;
				++cnt;
				swap(A[cnt],A[j]);
				for (j=cnt+1;j<=m;++j)
					if (A[j][i]) A[j]^=A[cnt]; 
		}
		return power(2,n-cnt);
}

int main()
{
		scanf("%d%d",&n,&m);
		for (i=1;i<=n;++i) 
			for (j=1;j<=m;++j) scanf("%d",&a[i][j]);
		for (i=1;i<=m;++i)
			for (j=1;j<=n;++j) A[i][j]=a[j][i];
		printf("%d\n",1ll*(power(2,n)-Guass()+Mo)%Mo*power(2,m-1)%Mo); 
}
### Prompt

Develop a solution in CPP to the problem described below:

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<cstring>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<ctime>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<bitset>

#define N 305
#define M 8000005

#define ls (t<<1)
#define rs ((t<<1)|1)
#define mid ((l+r)>>1)

#define mk make_pair
#define pb push_back
#define fi first
#define se second

#define Mo 998244353

using namespace std;

bitset<305>A[305];

int a[N][N],n,m,i,j;

int power(int x,int y)
{
		int s=1;
		for(;y;y>>=1)
		{
				if(y&1) s=1ll*s*x%Mo;
				x=1ll*x*x%Mo; 
		}
		return s;
}

int Guass()
{
		int i,j,k,cnt=0;
		for (i=1;i<=n;++i)
		{
				for (j=cnt+1;j<=m;++j) if (A[j][i]) break;
				if (j>m) continue;
				++cnt;
				swap(A[cnt],A[j]);
				for (j=cnt+1;j<=m;++j)
					if (A[j][i]) A[j]^=A[cnt]; 
		}
		return power(2,n-cnt);
}

int main()
{
		scanf("%d%d",&n,&m);
		for (i=1;i<=n;++i) 
			for (j=1;j<=m;++j) scanf("%d",&a[i][j]);
		for (i=1;i<=m;++i)
			for (j=1;j<=n;++j) A[i][j]=a[j][i];
		printf("%d\n",1ll*(power(2,n)-Guass()+Mo)%Mo*power(2,m-1)%Mo); 
}
```