#include<bits/stdc++.h>
#define ll long long
#define mo 998244353
#define N 310
using namespace std;
template <typename T> void read(T &x){
	x=0;char c=getchar();int fh=1;
	while (!isdigit(c)){if (c=='-')fh=-1;c=getchar();}
	while (isdigit(c))x=x*10+c-'0',c=getchar();
	x*=fh;
}
ll a[N][N],p[N][N];
ll m,n,sum;
ll po(ll x,ll y){ll z=1;while (y){if (y%2)z=x*z%mo;x=x*x%mo;y/=2;}return z;}
int main(){
	read(n);read(m);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)read(a[i][j]);
	for (int i=1;i<=n;i++){
		for (int j=m;j>=1;j--){
			if (p[j][j]==0){
				if (a[i][j]==1){
					for (int k=1;k<=j;k++) p[j][k]=a[i][k];
					sum++;
					break;
				}
			}else{
				if (a[i][j]==1)for (int k=1;k<=j;k++) a[i][k]^=p[j][k];
			}
		}
	}
	cout<<((po(2,n)-po(2,n-sum)+mo)%mo)*po(2,m-1)%mo<<endl;
	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>
#define ll long long
#define mo 998244353
#define N 310
using namespace std;
template <typename T> void read(T &x){
	x=0;char c=getchar();int fh=1;
	while (!isdigit(c)){if (c=='-')fh=-1;c=getchar();}
	while (isdigit(c))x=x*10+c-'0',c=getchar();
	x*=fh;
}
ll a[N][N],p[N][N];
ll m,n,sum;
ll po(ll x,ll y){ll z=1;while (y){if (y%2)z=x*z%mo;x=x*x%mo;y/=2;}return z;}
int main(){
	read(n);read(m);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)read(a[i][j]);
	for (int i=1;i<=n;i++){
		for (int j=m;j>=1;j--){
			if (p[j][j]==0){
				if (a[i][j]==1){
					for (int k=1;k<=j;k++) p[j][k]=a[i][k];
					sum++;
					break;
				}
			}else{
				if (a[i][j]==1)for (int k=1;k<=j;k++) a[i][k]^=p[j][k];
			}
		}
	}
	cout<<((po(2,n)-po(2,n-sum)+mo)%mo)*po(2,m-1)%mo<<endl;
	return 0;
}
```