Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

2 2
0 1
1 0

Sample Output 1

6

For example, if A consists of the first row and B consists of both columns, the sum of the numbers contained in the intersection is 0+1=1.

Sample Input 2

2 3
0 0 0
0 1 0

Sample Output 2

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
// define
#define int long long
#define UNIQUE(v) v.erase(unique(all(v)), v.end());
#define ZIP(v) sort(all(v)),UNIQUE(v)
#define ADD(a, b) a = (a + b) % mod
#define SUB(a, b) a = (a+mod-b)%mod
#define MUL(a, b) a = (a * b) % mod
#define rollcall cout << "I'm Sucu." << endl;
#define repi(i,m,n) for(int i = m;i < n;i++)
#define drep(i,n,m) for(int i = n;i >= m;i--)
#define rep(i,n) repi(i,0,n)
#define rrep(i,n) repi(i,1,n+1)
#define chmin(x,y) x = min(x,y)
#define chmax(x,y) x = max(x,y)
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(), v.rend()
#define dmp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define pf(x) push_front(x)
#define fi first
#define se second
// debug
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p){
  os << "(" << p.first << "," << p.second << ")";return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v){
  for (auto it = v.begin();it != v.end();++it){
    if(it != v.begin())os << " ";os << *it;
      }return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &mp){
  for(auto x: mp)os << "(" << x.first << "," << x.second << ")" << endl;
  return os;
}
template<typename T, int SIZE>
int array_length(const T (&)[SIZE]){return SIZE;}
template<typename T, int N>
void PRINTF(const T (&a)[N], int s = N, int t = -1, bool f = true){
	if(t == -1){rep(i,s){if(i)cout << " ";cout << a[i];}}
	else repi(i,s,t){if(i!=s)cout << " ";cout << a[i];}
	if(f)cout << "\n";
}
template<typename T, int N1, int N2>
void PRINTF(const T (&a)[N1][N2], int h = N1, int w = N2){
	rep(i,h){rep(j,w){cout << a[i][j] << " \n"[j==w-1];}}
}
string substr(const string &str, int S1, int S2 = -1){
  if(S2 == -1)return str.substr(S1);
  return str.substr(S1, S2-S1);
}
// typedef
typedef complex<double> Point;
typedef pair<int, int> P;
typedef pair<int, P> PP;
typedef pair<P, int> Pi;
typedef vector<int> vi;
typedef deque<int> dq;
const int inf = 1e9+7;
const int INF = 1e18+7;

const int mod = 998244353;

vector<int> mat[400];
void XOR(vi &A, vi &B){
  int n = A.size();
  rep(i,n)A[i] ^= B[i];
}

int Pow(int x, int n){
  if(n < 0)n += mod-1;
  int ans = 1;
  while(n > 0){
    if(n & 1)ans = (ans*x)%mod;
    x = (x*x)%mod;
    n >>= 1;
  }
  return ans;
}

int calc_rank(int h, int w){
  int r = 0;
  rep(j,w){
    int pivot = -1;
    rep(i,h)if(mat[i][j]){
      pivot = i;break;
    }
    if(pivot == -1)continue;
    swap(mat[pivot], mat[r]);
    rep(i,h)if(i != r){
      if(mat[i][j])XOR(mat[i], mat[r]);
    }
    r++;
    if(r >= h)break;
  }
  return r;
}

signed main(){
  int h, w;
  scanf("%lld%lld", &h, &w);
  rep(i,h)rep(j,w){
    int tmp;scanf("%lld", &tmp);
    mat[i].pb(tmp);
  }
  int r = calc_rank(h, w);
  int ans = mod+Pow(2, h)-Pow(2, h-r);
  ans = (ans%mod)*Pow(2, w-1);
  printf("%lld\n", ans%mod);
  return 0;
}