#include<bits/stdc++.h>
using namespace std;
const int p=998244353;
int n,m,a[505][505],pw[505];
bitset<505> b[505];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		bitset<505> x;
		for(int j=1,y;j<=m;j++){
			scanf("%d",&y);
			x[j-1]=y;
		}
		for(int j=m-1;j>=0;j--)
			if(x[j]){
				if(b[j].any())x^=b[j];
				else{b[j]=x;break;}
			}
	}
	for(int i=pw[0]=1;i<=n||i<=m;i++)pw[i]=2*pw[i-1]%p;
	int rk=0;
	for(int i=0;i<=m-1;i++)if(b[i].any())rk++;
	int ans=1ll*(pw[n]-pw[n-rk])*pw[m-1]%p;
	printf("%d\n",(ans+p)%p);
	return 0;
}

### Prompt

Please provide a Cpp coded solution 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<bits/stdc++.h>
using namespace std;
const int p=998244353;
int n,m,a[505][505],pw[505];
bitset<505> b[505];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		bitset<505> x;
		for(int j=1,y;j<=m;j++){
			scanf("%d",&y);
			x[j-1]=y;
		}
		for(int j=m-1;j>=0;j--)
			if(x[j]){
				if(b[j].any())x^=b[j];
				else{b[j]=x;break;}
			}
	}
	for(int i=pw[0]=1;i<=n||i<=m;i++)pw[i]=2*pw[i-1]%p;
	int rk=0;
	for(int i=0;i<=m-1;i++)if(b[i].any())rk++;
	int ans=1ll*(pw[n]-pw[n-rk])*pw[m-1]%p;
	printf("%d\n",(ans+p)%p);
	return 0;
}

```