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;

size_t rankF2(vector<vector<long>>& a){ // calculate rank of matrix in F_2
    long n = a.size(), m = a[0].size();
    auto add = [&m](vector<long>& to, vector<long>& from){
        for(int i = 0; i < m; ++i)to[i] ^= from[i];
    };
    size_t now(0), ret(0);
    while(now < m && ret < n){
        size_t i(ret);
        for(; i < n; ++i)if(a[i][now])break;
        if(i == n){
            ++now;
            continue;
        }
        size_t j(i + 1);
        for(; j < n; ++j)add(a[j], a[i]);
        swap(a[ret], a[i]);
        ++ret;
    }
    return ret;
}

int main(){
    constexpr long m = 998244353;
    auto modpow = [&m](long n){
        long r = 1, p = 2;
        while(n){
            if(n & 1){
                r *= p;
                r %= m;
            }
            p *= p;
            p %= m;
            n >>= 1;
        }
        return r;
    };

    long N, M;
    cin >> N >> M;
    bool f = false;
    if(N > M && (f = true))swap(N, M);

    vector<vector<long>> a(N, vector<long>(M));
    if(f)for(int i = 0; i < M; ++i)for(int j = 0; j < N; ++j)scanf("%ld", &a[j][i]);
    else for(auto& i : a)for(auto &j : i)scanf("%ld", &j);

    auto r = rankF2(a);
    cout << modpow(N + M - r - 1) * (modpow(r) - 1) % m << endl;
}