#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<bitset>
using namespace std;
#define MAXN 300
#define MOD 998244353
int n,m;
bitset<MAXN+5> A[MAXN+5],B[MAXN+5];
int read()
{
	int x=0,f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f; 
}
int fst_pow(int a,int b)
{
	int res=1;
	while(b)
	{
		if(b&1)res=(1LL*res*a)%MOD;
		a=(1LL*a*a)%MOD;
		b>>=1;
	}
	return res;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		A[i][j]=(read()==1);
	int r=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		if(A[i][j])
		{
			if(B[j][j])A[i]^=B[j];
			else
			{B[j]=A[i];r++;break;}
		}
	printf("%d",(fst_pow(2,n+m-1)-fst_pow(2,n-r+m-1)+MOD)%MOD);
}
### 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<bitset>
using namespace std;
#define MAXN 300
#define MOD 998244353
int n,m;
bitset<MAXN+5> A[MAXN+5],B[MAXN+5];
int read()
{
	int x=0,f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f; 
}
int fst_pow(int a,int b)
{
	int res=1;
	while(b)
	{
		if(b&1)res=(1LL*res*a)%MOD;
		a=(1LL*a*a)%MOD;
		b>>=1;
	}
	return res;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		A[i][j]=(read()==1);
	int r=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		if(A[i][j])
		{
			if(B[j][j])A[i]^=B[j];
			else
			{B[j]=A[i];r++;break;}
		}
	printf("%d",(fst_pow(2,n+m-1)-fst_pow(2,n-r+m-1)+MOD)%MOD);
}
```