#include "bits/stdc++.h"

using namespace std;

constexpr double eps = 1e-9;
using matrix = vector<vector<int>>;

int myrank(matrix A) {
  const int n = A.size(), m = A[0].size();
  int r = 0;
  for (int i = 0; r < n && i < m; ++i) {
    int pivot = r;
    for (int j = r + 1; j < n; ++j)
      if (abs(A[j][i]) > abs(A[pivot][i])) pivot = j;
    swap(A[pivot], A[r]);
    if (abs(A[r][i]) < eps) continue;
    for (int k = m - 1; k >= i; --k) A[r][k] /= A[r][i];
    for (int j = r + 1; j < n; ++j)
      for (int k = i; k < m; ++k) A[j][k] -= A[r][k] * A[j][i];
    ++r;
  }
  return r;
}

using ll = long long;
constexpr ll mod = 998244353;

template <typename T>
T mypow(T a, T b) {
  if (b == 0) return 1;
  T tmp = mypow(a, b / 2);
  if (b % 2)
    return (((tmp * tmp) % mod) * a) % mod;
  else
    return (tmp * tmp) % mod;
}

int main() {
  ll n, m;
  cin >> n >> m;
  matrix a(n, vector<int>(m));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }

  int rank = 0;
  for (int j = 0; j < m; j++) {
    int pivot = -1;
    for (int i = rank; i < n; i++) {
      if (a[i][j]) {
        pivot = i;
        break;
      }
    }
    if (pivot >= 0) {
      if (pivot != rank)
        for (int k = 0; k < m; k++) swap(a[pivot][k], a[rank][k]);
      for (int i = rank + 1; i < n; i++) {
        if (a[i][j])
          for (int k = 0; k < m; k++) a[i][k] ^= a[rank][k];
      }
      rank++;
    }
  }

  ll ret = mypow(2LL, n + m - 1) + mod - mypow(2LL, n + m - rank - 1);
  cout << ret % mod << endl;
}

### Prompt

Please create a solution in Cpp 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;

constexpr double eps = 1e-9;
using matrix = vector<vector<int>>;

int myrank(matrix A) {
  const int n = A.size(), m = A[0].size();
  int r = 0;
  for (int i = 0; r < n && i < m; ++i) {
    int pivot = r;
    for (int j = r + 1; j < n; ++j)
      if (abs(A[j][i]) > abs(A[pivot][i])) pivot = j;
    swap(A[pivot], A[r]);
    if (abs(A[r][i]) < eps) continue;
    for (int k = m - 1; k >= i; --k) A[r][k] /= A[r][i];
    for (int j = r + 1; j < n; ++j)
      for (int k = i; k < m; ++k) A[j][k] -= A[r][k] * A[j][i];
    ++r;
  }
  return r;
}

using ll = long long;
constexpr ll mod = 998244353;

template <typename T>
T mypow(T a, T b) {
  if (b == 0) return 1;
  T tmp = mypow(a, b / 2);
  if (b % 2)
    return (((tmp * tmp) % mod) * a) % mod;
  else
    return (tmp * tmp) % mod;
}

int main() {
  ll n, m;
  cin >> n >> m;
  matrix a(n, vector<int>(m));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }

  int rank = 0;
  for (int j = 0; j < m; j++) {
    int pivot = -1;
    for (int i = rank; i < n; i++) {
      if (a[i][j]) {
        pivot = i;
        break;
      }
    }
    if (pivot >= 0) {
      if (pivot != rank)
        for (int k = 0; k < m; k++) swap(a[pivot][k], a[rank][k]);
      for (int i = rank + 1; i < n; i++) {
        if (a[i][j])
          for (int k = 0; k < m; k++) a[i][k] ^= a[rank][k];
      }
      rank++;
    }
  }

  ll ret = mypow(2LL, n + m - 1) + mod - mypow(2LL, n + m - rank - 1);
  cout << ret % mod << endl;
}

```