#include<bitset>
#include<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
using namespace std;

const int N=510,mod=998244353;
int n,m,y,rk,ans,a[N][N],pw[N];
bitset<N> b[N];

int main(){
	scanf("%d%d",&n,&m); pw[0]=1;
	rep(i,1,n){
		bitset<N>x(0);
		rep(j,1,m) scanf("%d",&y),x[j-1]=y;
		for(int j=m-1; ~j; j--)
			if (x[j]){
				if (b[j].any()) x^=b[j];
				 else{ b[j]=x; break; }
			}
	}
	rep(i,1,max(n,m)) pw[i]=2*pw[i-1]%mod;
	rep(i,0,m-1) if (b[i].any()) rk++;
	ans=1ll*(pw[n]-pw[n-rk])*pw[m-1]%mod;
	printf("%d\n",(ans+mod)%mod);
	return 0;
}
### Prompt

Please formulate 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<bitset>
#include<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
using namespace std;

const int N=510,mod=998244353;
int n,m,y,rk,ans,a[N][N],pw[N];
bitset<N> b[N];

int main(){
	scanf("%d%d",&n,&m); pw[0]=1;
	rep(i,1,n){
		bitset<N>x(0);
		rep(j,1,m) scanf("%d",&y),x[j-1]=y;
		for(int j=m-1; ~j; j--)
			if (x[j]){
				if (b[j].any()) x^=b[j];
				 else{ b[j]=x; break; }
			}
	}
	rep(i,1,max(n,m)) pw[i]=2*pw[i-1]%mod;
	rep(i,0,m-1) if (b[i].any()) rk++;
	ans=1ll*(pw[n]-pw[n-rk])*pw[m-1]%mod;
	printf("%d\n",(ans+mod)%mod);
	return 0;
}
```