#include<cstdio>
#include<iostream>
using namespace std;
const int N=3e2+10;
const int mod=998244353;
int n,m,a[N][N],c[N][N],b[N],Max=N-5,cnt;
inline int fas(int x,int p){int res=1;while(p){if(p&1)res=1ll*res*x%mod;p>>=1;x=1ll*x*x%mod;}return res;}
int main(){
	scanf("%d%d",&n,&m);
	for(register int i=1;i<=n;i++)
		for(register int j=1;j<=m;j++)
			scanf("%d",&a[i][j]);
	for(register int i=1;i<=n;i++){
		for(register int j=1;j<=m;j++)b[j]=a[i][j];
		for(register int p=Max;p;p--){
			if(b[p]){
				if(!c[p][p]){
					for(register int k=p;k;k--)c[p][k]=b[k];
					cnt++;break;
				}
				else for(register int k=p;k;k--)b[k]^=c[p][k];
			}
		}
	}
	printf("%lld\n",1ll*(fas(2,n)-fas(2,n-cnt)+mod)%mod*fas(2,m-1)%mod);
	return 0;
}
### Prompt

Your challenge is to write 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<cstdio>
#include<iostream>
using namespace std;
const int N=3e2+10;
const int mod=998244353;
int n,m,a[N][N],c[N][N],b[N],Max=N-5,cnt;
inline int fas(int x,int p){int res=1;while(p){if(p&1)res=1ll*res*x%mod;p>>=1;x=1ll*x*x%mod;}return res;}
int main(){
	scanf("%d%d",&n,&m);
	for(register int i=1;i<=n;i++)
		for(register int j=1;j<=m;j++)
			scanf("%d",&a[i][j]);
	for(register int i=1;i<=n;i++){
		for(register int j=1;j<=m;j++)b[j]=a[i][j];
		for(register int p=Max;p;p--){
			if(b[p]){
				if(!c[p][p]){
					for(register int k=p;k;k--)c[p][k]=b[k];
					cnt++;break;
				}
				else for(register int k=p;k;k--)b[k]^=c[p][k];
			}
		}
	}
	printf("%lld\n",1ll*(fas(2,n)-fas(2,n-cnt)+mod)%mod*fas(2,m-1)%mod);
	return 0;
}
```