#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<bitset>
#include<stack>
using namespace std;
#define MOD 998244353
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

bitset<324> mat[324];
Int cnt,n, m,a;
Int mod_pow(Int x, Int a, Int m = MOD){
  if(a == 0)return 1;
  Int res = mod_pow(x, a / 2, m);
  res = res * res % m;
  if(a % 2)res *= x;
  return res % m;
}

Int inv(Int x, Int m = MOD){
  return mod_pow(x, m-2, m);
}


int main(){
  cin >> n >> m;
  for(int i = 0;i < n;i++){
    for(int j = 0;j < m;j++){
      cin >> a;
      if(a)mat[i].set(j);
    }
  }
  int p = 0;
  for(int i = 0;i < n;i++){
    if(p == m)continue;
    int st = i;
    while(st < n && !mat[st][p])st++;
    if(st == n){
      p++;
      i--;
      continue;
    }
    cnt++;
    if(st != i)swap(mat[st], mat[i]);
    for(int j = i+1;j < n;j++){
      if(mat[j][p])mat[j] ^= mat[i];
    }
  }
  cout << ((mod_pow(2, n+m) - mod_pow(2,n+m-cnt))*inv(2) % MOD+ MOD )% MOD << endl;
  return 0;
}

### Prompt

Please formulate a CPP solution 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<bitset>
#include<stack>
using namespace std;
#define MOD 998244353
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

bitset<324> mat[324];
Int cnt,n, m,a;
Int mod_pow(Int x, Int a, Int m = MOD){
  if(a == 0)return 1;
  Int res = mod_pow(x, a / 2, m);
  res = res * res % m;
  if(a % 2)res *= x;
  return res % m;
}

Int inv(Int x, Int m = MOD){
  return mod_pow(x, m-2, m);
}


int main(){
  cin >> n >> m;
  for(int i = 0;i < n;i++){
    for(int j = 0;j < m;j++){
      cin >> a;
      if(a)mat[i].set(j);
    }
  }
  int p = 0;
  for(int i = 0;i < n;i++){
    if(p == m)continue;
    int st = i;
    while(st < n && !mat[st][p])st++;
    if(st == n){
      p++;
      i--;
      continue;
    }
    cnt++;
    if(st != i)swap(mat[st], mat[i]);
    for(int j = i+1;j < n;j++){
      if(mat[j][p])mat[j] ^= mat[i];
    }
  }
  cout << ((mod_pow(2, n+m) - mod_pow(2,n+m-cnt))*inv(2) % MOD+ MOD )% MOD << endl;
  return 0;
}

```