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

typedef long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}

ll nextLong() { ll x; scanf("%lld", &x); return x;}

const ll MOD = 998244353;


// 01ベクトルのランクを求める
int binary_matrix_rank(vector<vector<int>> A) {
  const int N = A.size();
  const int M = A[0].size();

  auto at = [](const vector<int> Ai, int j) -> int {
    return Ai[j];
  };

  int s = 0;
  REP(j, M) {
    int p = -1;
    for (int i = s; i < N; i++) if (at(A[i], j)) { p = i; break; }
    if (p >= 0) {
      swap(A[s], A[p]);
      for (int i = s+1; i < N; i++) if (at(A[i], j)) {
        REP(k, M)
        A[i][k] ^= A[s][k];
      }
      s++;
    }
  }
  return s;
}


int main2() {
  int N = nextLong();
  int M = nextLong();

  vector< vector<int> > A(N, vector<int>(M, 0));
  REP(i, N) REP(j, M) A[i][j] = nextLong();

  int R = binary_matrix_rank(A);

  ll ans = 1;
  REP(i, R) ans = (ans * 2) % MOD;
  ans = (ans - 1 + MOD) % MOD;

  REP(i, N-R) ans = (ans * 2) % MOD;
  REP(i, M-1) ans = (ans * 2) % MOD;

  cout << ans << endl;
  return 0;
}

int main() {

#ifdef LOCAL
  for (;!cin.eof();cin>>ws)
#endif
    main2();
  return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}

ll nextLong() { ll x; scanf("%lld", &x); return x;}

const ll MOD = 998244353;


// 01ベクトルのランクを求める
int binary_matrix_rank(vector<vector<int>> A) {
  const int N = A.size();
  const int M = A[0].size();

  auto at = [](const vector<int> Ai, int j) -> int {
    return Ai[j];
  };

  int s = 0;
  REP(j, M) {
    int p = -1;
    for (int i = s; i < N; i++) if (at(A[i], j)) { p = i; break; }
    if (p >= 0) {
      swap(A[s], A[p]);
      for (int i = s+1; i < N; i++) if (at(A[i], j)) {
        REP(k, M)
        A[i][k] ^= A[s][k];
      }
      s++;
    }
  }
  return s;
}


int main2() {
  int N = nextLong();
  int M = nextLong();

  vector< vector<int> > A(N, vector<int>(M, 0));
  REP(i, N) REP(j, M) A[i][j] = nextLong();

  int R = binary_matrix_rank(A);

  ll ans = 1;
  REP(i, R) ans = (ans * 2) % MOD;
  ans = (ans - 1 + MOD) % MOD;

  REP(i, N-R) ans = (ans * 2) % MOD;
  REP(i, M-1) ans = (ans * 2) % MOD;

  cout << ans << endl;
  return 0;
}

int main() {

#ifdef LOCAL
  for (;!cin.eof();cin>>ws)
#endif
    main2();
  return 0;
}
```