#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll N=320,mo=998244353;

int a[N][N];
int n,m;

inline ll ksm(ll o,ll t)
{
	ll y=1;
	for(;t;t>>=1,o=o*o%mo)
	if(t&1)y=y*o%mo;
	return y;
}

bool bz[N];

inline int get()
{
	fo(i,1,m){
		int kk=0;
		fo(l,1,n)if(a[l][i]==1&&!bz[l]){
			kk=l; break;
		}
		bz[kk]=true;
		fo(l,1,n)if(!bz[l]&&a[l][i])
		fo(j,i,m)a[l][j]^=a[kk][j];
	}
	int ans=0;
	fo(i,1,n){
		int op=0;
		fo(l,1,m)op=op|a[i][l];
		ans=ans+op;
	}
	return ans;
}

int main()
{
	scanf("%d%d",&n,&m);
	fo(i,1,n)fo(l,1,m)scanf("%d",&a[i][l]);
	int r=get();
	ll da=(ksm(2,n+m-1)-ksm(2,n+m-1-r)+mo)%mo;
	cout<<da;
}
### 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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll N=320,mo=998244353;

int a[N][N];
int n,m;

inline ll ksm(ll o,ll t)
{
	ll y=1;
	for(;t;t>>=1,o=o*o%mo)
	if(t&1)y=y*o%mo;
	return y;
}

bool bz[N];

inline int get()
{
	fo(i,1,m){
		int kk=0;
		fo(l,1,n)if(a[l][i]==1&&!bz[l]){
			kk=l; break;
		}
		bz[kk]=true;
		fo(l,1,n)if(!bz[l]&&a[l][i])
		fo(j,i,m)a[l][j]^=a[kk][j];
	}
	int ans=0;
	fo(i,1,n){
		int op=0;
		fo(l,1,m)op=op|a[i][l];
		ans=ans+op;
	}
	return ans;
}

int main()
{
	scanf("%d%d",&n,&m);
	fo(i,1,n)fo(l,1,m)scanf("%d",&a[i][l]);
	int r=get();
	ll da=(ksm(2,n+m-1)-ksm(2,n+m-1-r)+mo)%mo;
	cout<<da;
}
```