#include <bits/stdc++.h>

using namespace std;

using int64 = long long;
const int mod = 998244353;


int main() {
  int N, M;
  bitset< 300 > X[300];
  cin >> N >> M;
  int rank = 0;
  for(int i = 0; i < N; i++) {
    bitset< 300 > tap;
    for(int j = 0; j < M; j++) {
      int p;
      cin >> p;
      tap[j] = p;
    }
    for(int j = 0; j < M; j++) {
      if(tap[j]) {
        tap ^= X[j];
        if(tap[j]) {
          X[j] = tap;
          ++rank;
          break;
        }
      }
    }
  }
  int64 all = 1, sub = 1;
  for(int i = 0; i < N + M - 1; i++) (all *= 2) %= mod;
  for(int i = 0; i < N + M - rank - 1; i++) (sub *= 2) %= mod;
  cout << (all + mod - sub) % mod << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;

using int64 = long long;
const int mod = 998244353;


int main() {
  int N, M;
  bitset< 300 > X[300];
  cin >> N >> M;
  int rank = 0;
  for(int i = 0; i < N; i++) {
    bitset< 300 > tap;
    for(int j = 0; j < M; j++) {
      int p;
      cin >> p;
      tap[j] = p;
    }
    for(int j = 0; j < M; j++) {
      if(tap[j]) {
        tap ^= X[j];
        if(tap[j]) {
          X[j] = tap;
          ++rank;
          break;
        }
      }
    }
  }
  int64 all = 1, sub = 1;
  for(int i = 0; i < N + M - 1; i++) (all *= 2) %= mod;
  for(int i = 0; i < N + M - rank - 1; i++) (sub *= 2) %= mod;
  cout << (all + mod - sub) % mod << endl;
}

```