#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
typedef unsigned int uint;
using namespace std;

const ll mod = 998244353LL;

ll power_mod(ll num, ll power) {
    ll prod = 1;
    num %= mod;
    while (power > 0) {
        if (power & 1) {
            prod = prod * num % mod;
        }
        num = num * num % mod;
        power >>= 1;
    }
    return prod;
}

int main() {
    cin.sync_with_stdio(false);
    cin.tie(0);

    int n, m, A[300][300];
    cin >> n >> 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++;
        }
    }

    cout << (power_mod(2LL, (ll)(n+m-1)) - power_mod(2LL, (ll)(n+m-rank-1)) + mod) % mod << "\n";
    return 0;
}
### 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 <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
typedef unsigned int uint;
using namespace std;

const ll mod = 998244353LL;

ll power_mod(ll num, ll power) {
    ll prod = 1;
    num %= mod;
    while (power > 0) {
        if (power & 1) {
            prod = prod * num % mod;
        }
        num = num * num % mod;
        power >>= 1;
    }
    return prod;
}

int main() {
    cin.sync_with_stdio(false);
    cin.tie(0);

    int n, m, A[300][300];
    cin >> n >> 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++;
        }
    }

    cout << (power_mod(2LL, (ll)(n+m-1)) - power_mod(2LL, (ll)(n+m-rank-1)) + mod) % mod << "\n";
    return 0;
}
```