#include<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_set>
#include<unordered_map>
#include<fstream>
#include<list>
#include<functional>
#include<bitset>
#include<complex>
#include<tuple>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pi;
typedef pair<double,double> pd;
typedef pair<double,ll> pdl;
#define F first
#define S second
const ll E=1e18+7;
const ll MOD=998244353; //1000000007;

ll mod_pow(ll a,ll x){
    ll ret=1;
    while(x>0){
        if(x&1){(ret*=a)%=MOD;}
        (a*=a)%=MOD;
        x>>=1;
    }
    return ret;
}

int main(){
    ll n,m;
    cin>>n>>m;
    vector<vector<ll>> A(n,vector<ll>(m));
    for(auto &I:A){for(auto &T:I){cin>>T;}}
    ll cnt=m;
    ll w=0;
    for(int i=0;i<n && w<m;i++,w++){
        for(int h=i+1;h<n;h++){
            if(A[i][w]==0 && A[h][w]==1){swap(A[i],A[h]);}
        }
        if(A[i][w]==0){cnt--; i--; continue;}
        for(int h=i+1;h<n;h++){
            if(A[h][w]){
                for(int w=0;w<m;w++){
                    A[h][w]^=A[i][w];
                }
            }
        }
        if(i+1==n){cnt-=m-w-1;}
    }
    ll ans=mod_pow(2,n+m-1)-mod_pow(2,n+m-cnt-1);
    ans%=MOD;
    if(ans<0){ans+=MOD;}
    cout<<ans<<endl;
    
    
    
    
    return 0;
}

### 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<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_set>
#include<unordered_map>
#include<fstream>
#include<list>
#include<functional>
#include<bitset>
#include<complex>
#include<tuple>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pi;
typedef pair<double,double> pd;
typedef pair<double,ll> pdl;
#define F first
#define S second
const ll E=1e18+7;
const ll MOD=998244353; //1000000007;

ll mod_pow(ll a,ll x){
    ll ret=1;
    while(x>0){
        if(x&1){(ret*=a)%=MOD;}
        (a*=a)%=MOD;
        x>>=1;
    }
    return ret;
}

int main(){
    ll n,m;
    cin>>n>>m;
    vector<vector<ll>> A(n,vector<ll>(m));
    for(auto &I:A){for(auto &T:I){cin>>T;}}
    ll cnt=m;
    ll w=0;
    for(int i=0;i<n && w<m;i++,w++){
        for(int h=i+1;h<n;h++){
            if(A[i][w]==0 && A[h][w]==1){swap(A[i],A[h]);}
        }
        if(A[i][w]==0){cnt--; i--; continue;}
        for(int h=i+1;h<n;h++){
            if(A[h][w]){
                for(int w=0;w<m;w++){
                    A[h][w]^=A[i][w];
                }
            }
        }
        if(i+1==n){cnt-=m-w-1;}
    }
    ll ans=mod_pow(2,n+m-1)-mod_pow(2,n+m-cnt-1);
    ans%=MOD;
    if(ans<0){ans+=MOD;}
    cout<<ans<<endl;
    
    
    
    
    return 0;
}

```