#include <bits/stdc++.h>
using namespace std;
const int M = 305, p = 998244353;
int main() {
  do {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  } while (false);
  int n, m;
  cin >> n >> m;
  vector<bitset<M> > arr(n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      int x;
      cin >> x;
      arr[i][j] = x;
    }
  }
  int rank = 0;
  for (int i = 0; i < m; i++) {
    int row = rank;
    while (row < n && !arr[row][i]) {
      row++;
    }
    if (row < n) {
      swap(arr[row], arr[rank]);
      for (int j = rank + 1; j < n; j++) {
        if (arr[j][i]) {
          arr[j] ^= arr[rank];
        }
      }
      rank++;
    }
  }
  vector<int> pow(n + m);
  pow[0] = 1;
  for (int i = 1; i < n + m; i++) {
    pow[i] = pow[i - 1] * 2 % p;
  }
  cout << (pow[n + m - 1] - pow[n - rank + m - 1] + p) % p << endl;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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;
const int M = 305, p = 998244353;
int main() {
  do {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  } while (false);
  int n, m;
  cin >> n >> m;
  vector<bitset<M> > arr(n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      int x;
      cin >> x;
      arr[i][j] = x;
    }
  }
  int rank = 0;
  for (int i = 0; i < m; i++) {
    int row = rank;
    while (row < n && !arr[row][i]) {
      row++;
    }
    if (row < n) {
      swap(arr[row], arr[rank]);
      for (int j = rank + 1; j < n; j++) {
        if (arr[j][i]) {
          arr[j] ^= arr[rank];
        }
      }
      rank++;
    }
  }
  vector<int> pow(n + m);
  pow[0] = 1;
  for (int i = 1; i < n + m; i++) {
    pow[i] = pow[i - 1] * 2 % p;
  }
  cout << (pow[n + m - 1] - pow[n - rank + m - 1] + p) % p << endl;
  return 0;
}

```