#include <algorithm>
#include <cstdio>
#include <bitset>

const int N = 300;
const int MOD = 998244353;

using Bitset = std::bitset<N>;

int n, m, two[N + 1];
Bitset mat[N];

int main()
{
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++ i) {
    for (int j = 0, a; j < m; ++ j) {
      scanf("%d", &a);
      if (a == 1) {
        mat[i].set(j);
      }
    }
  }
  int rank = 0;
  for (int j = 0; j < m; ++ j) {
    int pivot = rank;
    while (pivot < n && !mat[pivot].test(j)) {
      pivot ++;
    }
    if (pivot == n) {
      continue;
    }
    std::swap(mat[rank], mat[pivot]);
    for (int i = rank + 1; i < n; ++ i) {
      if (mat[i].test(j)) {
        mat[i] ^= mat[rank];
      }
    }
    rank ++;
  }
  two[0] = 1;
  for (int i = 1; i <= N; ++ i) {
    two[i] = two[i - 1] * 2 % MOD;
  }
  int result = 1LL * (two[n] + MOD - two[n - rank]) * two[m - 1] % MOD;
  printf("%d\n", result);
}

### 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 <algorithm>
#include <cstdio>
#include <bitset>

const int N = 300;
const int MOD = 998244353;

using Bitset = std::bitset<N>;

int n, m, two[N + 1];
Bitset mat[N];

int main()
{
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++ i) {
    for (int j = 0, a; j < m; ++ j) {
      scanf("%d", &a);
      if (a == 1) {
        mat[i].set(j);
      }
    }
  }
  int rank = 0;
  for (int j = 0; j < m; ++ j) {
    int pivot = rank;
    while (pivot < n && !mat[pivot].test(j)) {
      pivot ++;
    }
    if (pivot == n) {
      continue;
    }
    std::swap(mat[rank], mat[pivot]);
    for (int i = rank + 1; i < n; ++ i) {
      if (mat[i].test(j)) {
        mat[i] ^= mat[rank];
      }
    }
    rank ++;
  }
  two[0] = 1;
  for (int i = 1; i <= N; ++ i) {
    two[i] = two[i - 1] * 2 % MOD;
  }
  int result = 1LL * (two[n] + MOD - two[n - rank]) * two[m - 1] % MOD;
  printf("%d\n", result);
}

```