#include<bits/stdc++.h>
#define cmin(a,b) (a>(b)?a=(b),1:0)
#define cmax(a,b) (a<(b)?a=(b),1:0)
#define dmin(a,b) ((a)<(b)?(a):(b))
#define dmax(a,b) ((a)>(b)?(a):(b))
namespace io
{
	int F()
	{
		 int F=1,n=0;
		 char ch;
		 while((ch=getchar())!='-'&&(ch<'0'||ch>'9'));
		 ch=='-'?F=0:n=ch-'0';
		 while((ch=getchar())>='0'&&ch<='9')n=n*10+ch-'0';
		 return F?n:-n;
	}
	long long G()
	{
		 long long F=1,n=0;
		 char ch;
		 while((ch=getchar())!='-'&&(ch<'0'||ch>'9'));
		 ch=='-'?F=0:n=ch-'0';
		 while((ch=getchar())>='0'&&ch<='9')n=n*10+ch-'0';
		 return F?n:-n;
	}
}
int R(int l,int r)
{
	return (rand()<<15|rand())%(r-l+1)+l;
}
std::bitset<333> a[333]; 
int ex[666];
const int M=998244353;
int main()
{
	int n=io::F(),m=io::F();
	for(register int i=1;i<=n;++i)
		for(register int j=1;j<=m;++j)
			a[i][j]=io::F();
	int up=1;
	for(register int i=1;i<=m;++i)
	{
		int pl=0;
		for(register int j=up;j<=n;++j)
			if(a[j].test(i)){pl=j;break;}
		if(pl==0)continue;
		if(pl!=up)std::swap(a[up],a[pl]);
		for(register int j=up+1;j<=n;++j)
			if(a[j].test(i))a[j]^=a[up];
		++up;
	}
	int cnt=n-up+1;
	ex[0]=1;
	for(register int i=1;i<=n+m;++i)ex[i]=ex[i-1]*2%M;
	printf("%lld\n",(ex[n]-ex[cnt]+M)*1ll*ex[m-1]%M);
	return 0;
}

### 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<bits/stdc++.h>
#define cmin(a,b) (a>(b)?a=(b),1:0)
#define cmax(a,b) (a<(b)?a=(b),1:0)
#define dmin(a,b) ((a)<(b)?(a):(b))
#define dmax(a,b) ((a)>(b)?(a):(b))
namespace io
{
	int F()
	{
		 int F=1,n=0;
		 char ch;
		 while((ch=getchar())!='-'&&(ch<'0'||ch>'9'));
		 ch=='-'?F=0:n=ch-'0';
		 while((ch=getchar())>='0'&&ch<='9')n=n*10+ch-'0';
		 return F?n:-n;
	}
	long long G()
	{
		 long long F=1,n=0;
		 char ch;
		 while((ch=getchar())!='-'&&(ch<'0'||ch>'9'));
		 ch=='-'?F=0:n=ch-'0';
		 while((ch=getchar())>='0'&&ch<='9')n=n*10+ch-'0';
		 return F?n:-n;
	}
}
int R(int l,int r)
{
	return (rand()<<15|rand())%(r-l+1)+l;
}
std::bitset<333> a[333]; 
int ex[666];
const int M=998244353;
int main()
{
	int n=io::F(),m=io::F();
	for(register int i=1;i<=n;++i)
		for(register int j=1;j<=m;++j)
			a[i][j]=io::F();
	int up=1;
	for(register int i=1;i<=m;++i)
	{
		int pl=0;
		for(register int j=up;j<=n;++j)
			if(a[j].test(i)){pl=j;break;}
		if(pl==0)continue;
		if(pl!=up)std::swap(a[up],a[pl]);
		for(register int j=up+1;j<=n;++j)
			if(a[j].test(i))a[j]^=a[up];
		++up;
	}
	int cnt=n-up+1;
	ex[0]=1;
	for(register int i=1;i<=n+m;++i)ex[i]=ex[i-1]*2%M;
	printf("%lld\n",(ex[n]-ex[cnt]+M)*1ll*ex[m-1]%M);
	return 0;
}

```