#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
template <class T>
inline void read(T &x)
{
    x=0; char c=getchar();  int f=1;
    while(!isdigit(c)){if(c=='-')f=-1; c=getchar();}
    while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
#include <bitset>
const int mod=998244353;
ll Pow(ll x,ll y){ll res=1; while(y){if(y&1) res=res*x%mod; x=x*x%mod,y>>=1;} return res;}
bitset<310> f[310];
int n,m;
void print()
{
	for(int i=0;i<n;++i,puts(""))
	{
		for(int j=0;j<m;++j) printf("%d ",f[i][j]==1);
	}
	puts("");
}
int main()
{
	int x; read(n),read(m);
	for(int i=0;i<n;++i) for(int j=0;j<m;++j) read(x),f[i][j]=x;
	int ans=0;
	for(int i=0;i<m;++i)
	{
		for(int j=0;j<n;++j) if((!f[j][j]||j>=i)&&f[j][i]){if(j!=i)swap(f[i],f[j]); break;}//必须循环到n 
		if(!f[i][i]){ans++; continue;}
		for(int j=0;j<n;++j) if(j!=i&&f[j][i]) f[j]^=f[i];
//		print(); 
	}
//	cout<<ans<<endl;
	printf("%lld",((Pow(2,m+n-1)-Pow(2,ans+n-1))%mod+mod)%mod);
	return 0;
}

### Prompt

Your task is to create 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>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
template <class T>
inline void read(T &x)
{
    x=0; char c=getchar();  int f=1;
    while(!isdigit(c)){if(c=='-')f=-1; c=getchar();}
    while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
#include <bitset>
const int mod=998244353;
ll Pow(ll x,ll y){ll res=1; while(y){if(y&1) res=res*x%mod; x=x*x%mod,y>>=1;} return res;}
bitset<310> f[310];
int n,m;
void print()
{
	for(int i=0;i<n;++i,puts(""))
	{
		for(int j=0;j<m;++j) printf("%d ",f[i][j]==1);
	}
	puts("");
}
int main()
{
	int x; read(n),read(m);
	for(int i=0;i<n;++i) for(int j=0;j<m;++j) read(x),f[i][j]=x;
	int ans=0;
	for(int i=0;i<m;++i)
	{
		for(int j=0;j<n;++j) if((!f[j][j]||j>=i)&&f[j][i]){if(j!=i)swap(f[i],f[j]); break;}//必须循环到n 
		if(!f[i][i]){ans++; continue;}
		for(int j=0;j<n;++j) if(j!=i&&f[j][i]) f[j]^=f[i];
//		print(); 
	}
//	cout<<ans<<endl;
	printf("%lld",((Pow(2,m+n-1)-Pow(2,ans+n-1))%mod+mod)%mod);
	return 0;
}

```