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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp51
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] dm = Console.ReadLine().Split().Select(x => int.Parse(x)).ToArray();
            int[][] mt = new int[dm[0]][];
            for (int i = 0; i < dm[0]; ++i)
                mt[i] = Console.ReadLine().Split().Select(x => int.Parse(x)).ToArray();
            Solution sol = new Solution();
            Console.WriteLine(sol.GetOddRectangles(mt, dm[0], dm[1]));
        }
    }


    public class Solution
    {
        public int GetOddRectangles(int[][] mt, int N, int M)
        {
            int[,] acc = new int[N + 1, M + 1];

            for (int i = 1; i <= N; ++i)
                for (int j = 1; j <= M; ++j)
                    acc[i, j] = acc[i - 1, j] + acc[i, j - 1] - acc[i - 1, j - 1] + mt[i - 1][j - 1];

            int sol = 0;
            for (int i = 1; i <= N; ++i)
                for (int j = 1; j <= M; ++j)
                    for (int l = i; l <= N; ++l)
                        for (int k = j; k <= M; ++k)
                            sol += (acc[l, k] - acc[l, j - 1] - acc[i - 1, k] + acc[i - 1, j - 1]) & 1;
            return sol % 998244353;
        }
    }
}