#include<bits/stdc++.h>
using namespace std;

const int maxN=310;
const int Mod=998244353;

int n,m;
bitset<maxN> B[maxN];
int C[maxN][maxN],Sm[maxN],pw[maxN];

int main(){
    pw[0]=1;for (int i=1;i<maxN;i++) pw[i]=2ll*pw[i-1]%Mod;
    scanf("%d%d",&n,&m);
    int cnt=0,one=0;
    for (int i=1;i<=n;i++){
        bitset<maxN> I;I.reset();
        for (int j=1;j<=m;j++){
            int x;scanf("%d",&x);I[j]=x;
        }
        bool flag=0;
        for (int j=1;j<=m;j++)
            if (I[j]){
                if (B[j]==0){
                    B[j]=I;flag=1;break;
                }
                else I^=B[j];
            }
        if (flag==0) ++cnt;
    }

    printf("%lld\n",1ll*(pw[n]-pw[cnt]+Mod)%Mod*pw[m-1]%Mod);return 0;
}
### Prompt

In CPP, your task is to solve 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;

const int maxN=310;
const int Mod=998244353;

int n,m;
bitset<maxN> B[maxN];
int C[maxN][maxN],Sm[maxN],pw[maxN];

int main(){
    pw[0]=1;for (int i=1;i<maxN;i++) pw[i]=2ll*pw[i-1]%Mod;
    scanf("%d%d",&n,&m);
    int cnt=0,one=0;
    for (int i=1;i<=n;i++){
        bitset<maxN> I;I.reset();
        for (int j=1;j<=m;j++){
            int x;scanf("%d",&x);I[j]=x;
        }
        bool flag=0;
        for (int j=1;j<=m;j++)
            if (I[j]){
                if (B[j]==0){
                    B[j]=I;flag=1;break;
                }
                else I^=B[j];
            }
        if (flag==0) ++cnt;
    }

    printf("%lld\n",1ll*(pw[n]-pw[cnt]+Mod)%Mod*pw[m-1]%Mod);return 0;
}
```