#include<bits/stdc++.h>
using namespace std;
typedef int ll;
typedef long double ld;
typedef pair<ll,ll> p2;
ll a[305][305],p[1005],hell=998244353;
vector<ll> basis[305];
ll insert(vector<ll> &v){
	ll n=ll(v.size());
	for(ll i=0;i<n;i++){
		if(!v[i])
		continue;
		if(basis[i].empty()){
			basis[i]=v;
			return 1;
		}
		for(ll j=0;j<n;j++)
		v[j]^=basis[i][j];
	}
	return 0;
}
void solve(){
	ll n,m,res=0;
	cin>>n>>m;
	for(ll i=0;i<n;i++)
	for(ll j=0;j<m;j++)
	cin>>a[i][j];
	for(ll j=0;j<m;j++){
		vector<ll> v;
		for(ll i=0;i<n;i++)
		v.push_back(a[i][j]);
		res+=insert(v);
	}
	ll ans=(p[n+m-1]-p[n+m-res-1])%hell;
	if(ans<0)
	ans+=hell;
	cout<<ans;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    p[0]=1;
    for(ll i=1;i<1005;i++)
    p[i]=(2*p[i-1])%hell;
    ll qc=1;
    //cin>>qc;
    for(ll i=1;i<=qc;i++)
	solve();
}

### 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>
using namespace std;
typedef int ll;
typedef long double ld;
typedef pair<ll,ll> p2;
ll a[305][305],p[1005],hell=998244353;
vector<ll> basis[305];
ll insert(vector<ll> &v){
	ll n=ll(v.size());
	for(ll i=0;i<n;i++){
		if(!v[i])
		continue;
		if(basis[i].empty()){
			basis[i]=v;
			return 1;
		}
		for(ll j=0;j<n;j++)
		v[j]^=basis[i][j];
	}
	return 0;
}
void solve(){
	ll n,m,res=0;
	cin>>n>>m;
	for(ll i=0;i<n;i++)
	for(ll j=0;j<m;j++)
	cin>>a[i][j];
	for(ll j=0;j<m;j++){
		vector<ll> v;
		for(ll i=0;i<n;i++)
		v.push_back(a[i][j]);
		res+=insert(v);
	}
	ll ans=(p[n+m-1]-p[n+m-res-1])%hell;
	if(ans<0)
	ans+=hell;
	cout<<ans;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    p[0]=1;
    for(ll i=1;i<1005;i++)
    p[i]=(2*p[i-1])%hell;
    ll qc=1;
    //cin>>qc;
    for(ll i=1;i<=qc;i++)
	solve();
}

```