#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <iostream>
#include <bitset>
using namespace std;
#define N 305
#define ll long long
#define mod 998244353
int n,m,b[N],ans=1;
bitset<N>a[N],t[N];
void insert(bitset<N> now)
{
	for(int i=m;i;i--)if(now[i])
		if(t[i][i])now^=t[i];
		else {t[i]=now;return ;}
	ans=ans<<1;ans%=mod;
}
int main()
{
	scanf("%d%d",&n,&m);int tt=1,pp=1;
	for(int i=1;i<m;i++)pp=(pp<<1)%mod;
	for(int i=1;i<=n;i++)
	{
		tt<<=1;tt%=mod;
		for(int j=1,x;j<=m;j++)
			scanf("%d",&x),a[i][j]=(x==1);
		insert(a[i]);
	}
	printf("%d\n",(ll)(tt-ans+mod)*pp%mod);
}
### Prompt

Generate 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 <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <iostream>
#include <bitset>
using namespace std;
#define N 305
#define ll long long
#define mod 998244353
int n,m,b[N],ans=1;
bitset<N>a[N],t[N];
void insert(bitset<N> now)
{
	for(int i=m;i;i--)if(now[i])
		if(t[i][i])now^=t[i];
		else {t[i]=now;return ;}
	ans=ans<<1;ans%=mod;
}
int main()
{
	scanf("%d%d",&n,&m);int tt=1,pp=1;
	for(int i=1;i<m;i++)pp=(pp<<1)%mod;
	for(int i=1;i<=n;i++)
	{
		tt<<=1;tt%=mod;
		for(int j=1,x;j<=m;j++)
			scanf("%d",&x),a[i][j]=(x==1);
		insert(a[i]);
	}
	printf("%d\n",(ll)(tt-ans+mod)*pp%mod);
}
```