#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 306;
const ll mod = 998244353;
typedef bitset<N> bst;

bst a[N],b[N],empty;

int n,m;

int cal() {
    int ret = 0;
    for (int i = 0; i < n; ++i)
        for (int j = m; j >= 0; --j)
            if (a[i][j]) {
                if (b[j] != empty) a[i] ^= b[j];
                else {
                    ++ret;
                    b[j] = a[i];
                    for (int k = j - 1; k >= 0; --k) if (b[k] != empty && (b[j][k])) b[j] ^= b[k];
                    for (int k = j + 1; k <= m; ++k) if (b[k][j]) b[k] ^= b[j];
                    break;
                }
            }
    return ret;
}

ll ppow(ll a,ll n) {
    ll ret=1,now=a;
    while (n) {
        if (n&1) {
            ret *= now;
            ret %= mod;
        }
        now *= now;
        now %= mod;
        n >>= 1;
    }
    return ret;
}

ll rev(ll x) {
    return ppow(x,mod-2);
}

int main () {
    //int n,m;
    cin >> n >> m;
    for (int i=0;i<n;++i) {
        for (int j=0;j<m;++j) {
            int x;
            cin >> x;
            a[i][j] = x;
        }
    }
    int x = cal();
    //cout << "x = " << x << endl;
    ll all_zero = ppow(2,n-x);
    //cout << "all_zero = " << all_zero << endl;
    ll ans = ppow(2,n+m);
    //cout << " ans = " << ans << endl;
    ans -= all_zero * ppow(2,m);
    //cout << "ans = " << ans << endl;
    ans = (ans%mod+mod)%mod;
    cout << ans * rev(2) % mod << endl;
}

### Prompt

Develop a solution in Cpp 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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 306;
const ll mod = 998244353;
typedef bitset<N> bst;

bst a[N],b[N],empty;

int n,m;

int cal() {
    int ret = 0;
    for (int i = 0; i < n; ++i)
        for (int j = m; j >= 0; --j)
            if (a[i][j]) {
                if (b[j] != empty) a[i] ^= b[j];
                else {
                    ++ret;
                    b[j] = a[i];
                    for (int k = j - 1; k >= 0; --k) if (b[k] != empty && (b[j][k])) b[j] ^= b[k];
                    for (int k = j + 1; k <= m; ++k) if (b[k][j]) b[k] ^= b[j];
                    break;
                }
            }
    return ret;
}

ll ppow(ll a,ll n) {
    ll ret=1,now=a;
    while (n) {
        if (n&1) {
            ret *= now;
            ret %= mod;
        }
        now *= now;
        now %= mod;
        n >>= 1;
    }
    return ret;
}

ll rev(ll x) {
    return ppow(x,mod-2);
}

int main () {
    //int n,m;
    cin >> n >> m;
    for (int i=0;i<n;++i) {
        for (int j=0;j<m;++j) {
            int x;
            cin >> x;
            a[i][j] = x;
        }
    }
    int x = cal();
    //cout << "x = " << x << endl;
    ll all_zero = ppow(2,n-x);
    //cout << "all_zero = " << all_zero << endl;
    ll ans = ppow(2,n+m);
    //cout << " ans = " << ans << endl;
    ans -= all_zero * ppow(2,m);
    //cout << "ans = " << ans << endl;
    ans = (ans%mod+mod)%mod;
    cout << ans * rev(2) % mod << endl;
}

```