#include <algorithm>
#include <bitset>
#include <cassert>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<unsigned char, unsigned char> pcc;
typedef pair<double, double> pdd;
typedef bitset<305> bs;

const int MOD = 998244353;

bs matrix[305];

ll modpow(ll b, ll e, ll m) {
  ll r = 1;
  while(e) {
    if(e%2) {
      r*=b;
      r%=m;
    }
    b*=b;
    b%=m;
    e/=2;
  }
  return r;
}

void solve() {
  int r, c;
  cin >> r >> c;
  for(int i = 0; i < r; i++) {
    for(int j = 0; j < c; j++) {
      int t;
      cin >> t;
      if(t) matrix[i].set(j);
    }
  }
  int rank = 0;
  for(int j = 0; j < c; j++) {
    bool found = false;
    for(int i = rank; i < r; i++) {
      if(matrix[i][j]) {
        found = true;
        if(i != rank) swap(matrix[i], matrix[rank]);
      }
    }
    if(found) {
      for(int i = rank+1; i < r; i++) {
        if(matrix[i][j]) matrix[i] ^= matrix[rank];
      }
      rank++;
    }
  }
  ll ret = modpow(2, r+c-1, MOD);
  ret -= modpow(2, r+c-1-rank, MOD);
  ret += MOD;
  cout << ret%MOD << "\n";
}

void casesolve() {
  int t;
  cin >> t;
  for(int i = 1; i <= t; i++) {
    cout << "Case #" << i << ":" << "\n";
    solve();
  }
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  solve();
  // casesolve();
}

### 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 <algorithm>
#include <bitset>
#include <cassert>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<unsigned char, unsigned char> pcc;
typedef pair<double, double> pdd;
typedef bitset<305> bs;

const int MOD = 998244353;

bs matrix[305];

ll modpow(ll b, ll e, ll m) {
  ll r = 1;
  while(e) {
    if(e%2) {
      r*=b;
      r%=m;
    }
    b*=b;
    b%=m;
    e/=2;
  }
  return r;
}

void solve() {
  int r, c;
  cin >> r >> c;
  for(int i = 0; i < r; i++) {
    for(int j = 0; j < c; j++) {
      int t;
      cin >> t;
      if(t) matrix[i].set(j);
    }
  }
  int rank = 0;
  for(int j = 0; j < c; j++) {
    bool found = false;
    for(int i = rank; i < r; i++) {
      if(matrix[i][j]) {
        found = true;
        if(i != rank) swap(matrix[i], matrix[rank]);
      }
    }
    if(found) {
      for(int i = rank+1; i < r; i++) {
        if(matrix[i][j]) matrix[i] ^= matrix[rank];
      }
      rank++;
    }
  }
  ll ret = modpow(2, r+c-1, MOD);
  ret -= modpow(2, r+c-1-rank, MOD);
  ret += MOD;
  cout << ret%MOD << "\n";
}

void casesolve() {
  int t;
  cin >> t;
  for(int i = 1; i <= t; i++) {
    cout << "Case #" << i << ":" << "\n";
    solve();
  }
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  solve();
  // casesolve();
}

```