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

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
const ll mod = 998244353;
ll N, M;
ll a[302][302];
ll beki(ll a, ll b) {
    if(b == 0) return 1;
    ll ans = beki(a, b / 2);
    ans = ans * ans % mod;
    if(b % 2 == 1) {
        ans = ans * a % mod;
    }
    return ans;
}

ll Rank(ll index) {
    if(index > min(N, M)) return 0;
    int nonzero_i = -1;
    int nonzero_j = -1;
    for(int i = index; i <= N; i++) {
        for(int j = index; j <= M; j++) {
            if(a[i][j] == 1) {
                nonzero_i = i;
                nonzero_j = j;
            }
        }
    }
    if(nonzero_i == -1) return 0;
    for(int i = index; i <= N; i++) {
        swap(a[i][nonzero_j], a[i][index]);
    }
    for(int j = index; j <= M; j++) {
        swap(a[nonzero_i][j], a[index][j]);
    }
    for(int i = index + 1; i <= N; i++) {
        if(a[i][index] == 0) continue;
        for(int j = index; j <= M; j++) {
            a[i][j] = (a[i][j] + a[index][j]) % 2;
        }
    }
    for(int j = index + 1; j <= M; j++) {
        a[index][j] = 0;
    }
    return Rank(index + 1) + 1;
}

int main() {
    //cout.precision(10);
    cin >> N >> M;
    for(int i = 1; i <= N; i++) {
        for(int j = 1; j <= M; j++) {
            cin >> a[i][j];
        }
    }
    ll R = Rank(1);
    ll ans = (beki(2, R) - 1) * (beki(2, R - 1)) % mod;
    ans = ans * (beki(2, N - R)) % mod;
    ans = ans * (beki(2, M - R)) % mod;
    cout << ans << endl;
    return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
const ll mod = 998244353;
ll N, M;
ll a[302][302];
ll beki(ll a, ll b) {
    if(b == 0) return 1;
    ll ans = beki(a, b / 2);
    ans = ans * ans % mod;
    if(b % 2 == 1) {
        ans = ans * a % mod;
    }
    return ans;
}

ll Rank(ll index) {
    if(index > min(N, M)) return 0;
    int nonzero_i = -1;
    int nonzero_j = -1;
    for(int i = index; i <= N; i++) {
        for(int j = index; j <= M; j++) {
            if(a[i][j] == 1) {
                nonzero_i = i;
                nonzero_j = j;
            }
        }
    }
    if(nonzero_i == -1) return 0;
    for(int i = index; i <= N; i++) {
        swap(a[i][nonzero_j], a[i][index]);
    }
    for(int j = index; j <= M; j++) {
        swap(a[nonzero_i][j], a[index][j]);
    }
    for(int i = index + 1; i <= N; i++) {
        if(a[i][index] == 0) continue;
        for(int j = index; j <= M; j++) {
            a[i][j] = (a[i][j] + a[index][j]) % 2;
        }
    }
    for(int j = index + 1; j <= M; j++) {
        a[index][j] = 0;
    }
    return Rank(index + 1) + 1;
}

int main() {
    //cout.precision(10);
    cin >> N >> M;
    for(int i = 1; i <= N; i++) {
        for(int j = 1; j <= M; j++) {
            cin >> a[i][j];
        }
    }
    ll R = Rank(1);
    ll ans = (beki(2, R) - 1) * (beki(2, R - 1)) % mod;
    ans = ans * (beki(2, N - R)) % mod;
    ans = ans * (beki(2, M - R)) % mod;
    cout << ans << endl;
    return 0;
}

```