#include<bits/stdc++.h>
using namespace std;
const int N=305;
const int mod=998244353;
int n,m,a[N][N],now[N],in[N],p[N][N],r;
inline int poww(int a,int b){
  int res=1;
  for(;b;b>>=1,a=1ll*a*a%mod)
    if(b&1)res=1ll*res*a%mod;
  return res;
}
int main()
{
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
  for(int i=1;i<=n;i++){
    memcpy(now,a[i],sizeof(now));
    for(int j=1;j<=m;j++)
      if(now[j]){
    if(in[j])for(int k=j;k<=m;k++)now[k]^=p[j][k];
    else{in[j]=1;memcpy(p[j],now,sizeof(p[j]));r++;break;}
      }
  }
  printf("%lld\n",1ll*(poww(2,n)-poww(2,n-r)+mod)%mod*poww(2,m-1)%mod);
  return 0;
}

### Prompt

Please create a solution in Cpp 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<bits/stdc++.h>
using namespace std;
const int N=305;
const int mod=998244353;
int n,m,a[N][N],now[N],in[N],p[N][N],r;
inline int poww(int a,int b){
  int res=1;
  for(;b;b>>=1,a=1ll*a*a%mod)
    if(b&1)res=1ll*res*a%mod;
  return res;
}
int main()
{
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
  for(int i=1;i<=n;i++){
    memcpy(now,a[i],sizeof(now));
    for(int j=1;j<=m;j++)
      if(now[j]){
    if(in[j])for(int k=j;k<=m;k++)now[k]^=p[j][k];
    else{in[j]=1;memcpy(p[j],now,sizeof(p[j]));r++;break;}
      }
  }
  printf("%lld\n",1ll*(poww(2,n)-poww(2,n-r)+mod)%mod*poww(2,m-1)%mod);
  return 0;
}

```