#include<bits/stdc++.h>
#define ll long long 
using namespace std;
const int _=305,yl=998244353;
int g[_],cnt,n,m;
bitset<_>a[_],f[_];
ll POW(ll x,ll y){
	ll res=1;
	while(y){
		if(y&1)res=res*x%yl;
		x=x*x%yl;y>>=1;
	}
	return res;
}
int main(){
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			int x;cin>>x;
			if(x)a[i].set(j);
		}
		for(int j=1;j<=m;++j){
			if(a[i][j]){
				if(g[j])a[i]^=f[j];
				else {f[j]=a[i];g[j]=1;++cnt;break;}
			}
		}
	}
	cout<<(POW(2,m)-POW(2,m-cnt)+yl)*POW(2,n-1)%yl<<endl;
}

### Prompt

Please create a solution in cpp 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<bits/stdc++.h>
#define ll long long 
using namespace std;
const int _=305,yl=998244353;
int g[_],cnt,n,m;
bitset<_>a[_],f[_];
ll POW(ll x,ll y){
	ll res=1;
	while(y){
		if(y&1)res=res*x%yl;
		x=x*x%yl;y>>=1;
	}
	return res;
}
int main(){
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			int x;cin>>x;
			if(x)a[i].set(j);
		}
		for(int j=1;j<=m;++j){
			if(a[i][j]){
				if(g[j])a[i]^=f[j];
				else {f[j]=a[i];g[j]=1;++cnt;break;}
			}
		}
	}
	cout<<(POW(2,m)-POW(2,m-cnt)+yl)*POW(2,n-1)%yl<<endl;
}

```