#include<bits/stdc++.h>

using namespace std;

using ll = long long;

const int N = 305, mod = 998244353;
int n, m, pow2[N << 1];
vector< bitset<N> > basis;

int lstBit(bitset<N> x) {
    int ret = -1;
    for (int i = 0; i < N; ++i) if (x[i]) ret = i;
    return ret;
}

void updBasis(bitset<N> val) {
    for (int i = 0; i < (int)basis.size(); ++i) {
        if (!val.count() ) return ;
        auto _ = basis[i];
        int lst_ = lstBit(_),
            lstVal = lstBit(val);
        if (lst_ == lstVal) val ^= _;
        else if (lst_ < lstVal) {
            basis.insert(basis.begin() + i, val);
            return ;
        }
    }
    if (val.count() ) basis.push_back(val);
}

int add(int _a, int _b) {
    _a += _b;
    if (_a >= mod) _a -= mod;
    return _a;
}
int sub(int _a, int _b) { return add(_a, mod - _b); }
int mul(int _a, int _b) { return (int)( (ll)_a * _b % mod); }

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        bitset<N> val;
        for (int j = 1; j <= m; ++j) {
            int a; cin >> a;
            val[j] = a;
        }
        updBasis(val);
    }

    pow2[0] = 1; for (int i = 1; i < (N << 1); ++i) pow2[i] = (int)(2 * pow2[i - 1] % mod);

    cout << sub(pow2[n + m - 1], pow2[n + m - 1 - (int)basis.size()]);

    return 0;
}

### 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 ll = long long;

const int N = 305, mod = 998244353;
int n, m, pow2[N << 1];
vector< bitset<N> > basis;

int lstBit(bitset<N> x) {
    int ret = -1;
    for (int i = 0; i < N; ++i) if (x[i]) ret = i;
    return ret;
}

void updBasis(bitset<N> val) {
    for (int i = 0; i < (int)basis.size(); ++i) {
        if (!val.count() ) return ;
        auto _ = basis[i];
        int lst_ = lstBit(_),
            lstVal = lstBit(val);
        if (lst_ == lstVal) val ^= _;
        else if (lst_ < lstVal) {
            basis.insert(basis.begin() + i, val);
            return ;
        }
    }
    if (val.count() ) basis.push_back(val);
}

int add(int _a, int _b) {
    _a += _b;
    if (_a >= mod) _a -= mod;
    return _a;
}
int sub(int _a, int _b) { return add(_a, mod - _b); }
int mul(int _a, int _b) { return (int)( (ll)_a * _b % mod); }

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        bitset<N> val;
        for (int j = 1; j <= m; ++j) {
            int a; cin >> a;
            val[j] = a;
        }
        updBasis(val);
    }

    pow2[0] = 1; for (int i = 1; i < (N << 1); ++i) pow2[i] = (int)(2 * pow2[i - 1] % mod);

    cout << sub(pow2[n + m - 1], pow2[n + m - 1 - (int)basis.size()]);

    return 0;
}

```