#include <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=998244353;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

const int M=305;
int n,m;
ll two[M];
vvi a;

int main(){
	two[0]++;
	for(int i=1;i<M;i++) two[i]=two[i-1]*2%mod;
	cin>>n>>m;
	a=vvi(n,vi(m));
	for(int i=0;i<n;i++) for(int j=0;j<m;j++) cin>>a[i][j];
	int I=0;
	for(int i=0;i<m;i++){
		for(int j=I;j<n;j++) if(a[j][i]){
			swap(a[I],a[j]);
			break;
		}
		if(I==n||!a[I][i]) continue;
		for(int j=0;j<n;j++) if(j!=I&&a[j][i]){
			for(int k=0;k<m;k++) a[j][k]^=a[I][k];
		}
		I++;
	}
	cout<<(two[m]-two[m-I]+mod)*two[n-1]%mod<<endl;
}
### Prompt

Create a solution in cpp for 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 syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=998244353;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

const int M=305;
int n,m;
ll two[M];
vvi a;

int main(){
	two[0]++;
	for(int i=1;i<M;i++) two[i]=two[i-1]*2%mod;
	cin>>n>>m;
	a=vvi(n,vi(m));
	for(int i=0;i<n;i++) for(int j=0;j<m;j++) cin>>a[i][j];
	int I=0;
	for(int i=0;i<m;i++){
		for(int j=I;j<n;j++) if(a[j][i]){
			swap(a[I],a[j]);
			break;
		}
		if(I==n||!a[I][i]) continue;
		for(int j=0;j<n;j++) if(j!=I&&a[j][i]){
			for(int k=0;k<m;k++) a[j][k]^=a[I][k];
		}
		I++;
	}
	cout<<(two[m]-two[m-I]+mod)*two[n-1]%mod<<endl;
}
```