#include<bits/stdc++.h>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define rep2(i,k,n) for(int i=k;i>=n;i--)
#define mod 998244353
using namespace std;
inline void upd(int& x,int y){
	x+=y;
	x-= (x>=mod) ? mod : 0;
}
const int N=305;
int pw[N<<1],f[N],a[N][N],n,m;
int main(){
	pw[0]=1;rep(i,1,600)pw[i]=pw[i-1],upd(pw[i],pw[i-1]);
	rep(i,1,N-1)f[i]=pw[(i-1)<<1],upd(f[i],f[i-1]),upd(f[i],f[i-1]);
	scanf("%d%d",&n,&m);
	rep(i,1,n)rep(j,1,m)scanf("%d",&a[i][j]);
	int r=0;
	rep(i,1,m){
		int t=0;
		rep(j,r+1,n)if(a[j][i]){t=j;break;}
		if(!t)continue;
		r++;
		rep(j,1,m)swap(a[r][j],a[t][j]);
		rep(j,r+1,n)if(a[j][i]){
			rep(k,i,m)a[j][k]^=a[r][k];
		}
	}
	int ans=1ll*f[r]*pw[n-r]%mod*pw[m-r]%mod;
	printf("%d\n",ans);
} 
### 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<bits/stdc++.h>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define rep2(i,k,n) for(int i=k;i>=n;i--)
#define mod 998244353
using namespace std;
inline void upd(int& x,int y){
	x+=y;
	x-= (x>=mod) ? mod : 0;
}
const int N=305;
int pw[N<<1],f[N],a[N][N],n,m;
int main(){
	pw[0]=1;rep(i,1,600)pw[i]=pw[i-1],upd(pw[i],pw[i-1]);
	rep(i,1,N-1)f[i]=pw[(i-1)<<1],upd(f[i],f[i-1]),upd(f[i],f[i-1]);
	scanf("%d%d",&n,&m);
	rep(i,1,n)rep(j,1,m)scanf("%d",&a[i][j]);
	int r=0;
	rep(i,1,m){
		int t=0;
		rep(j,r+1,n)if(a[j][i]){t=j;break;}
		if(!t)continue;
		r++;
		rep(j,1,m)swap(a[r][j],a[t][j]);
		rep(j,r+1,n)if(a[j][i]){
			rep(k,i,m)a[j][k]^=a[r][k];
		}
	}
	int ans=1ll*f[r]*pw[n-r]%mod*pw[m-r]%mod;
	printf("%d\n",ans);
} 
```