#include <bits/stdc++.h>
using namespace std;
#define finish(x) return cout << x << endl, 0
#define ll long long

const int N = 301;

int n, m, pow2[N], mod = 998244353;
vector <bitset <N> > b;

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    pow2[0] = 1;
    for(int i = 1 ; i < N ; i++){
        pow2[i] = 2 * pow2[i - 1] % mod;
    }
    cin >> n >> m;
    b.resize(n);
    for(int i = 0 ; i < n ; i++){
        for(int j = 0 ; j < m ; j++){
            int x;
            cin >> x;
            if(x) b[i].set(j);
        }
    }
    vector <int> vis(n);
    int h = n;
    for(int j = 0 ; j < m ; j++){
        int idx = -1;
        for(int i = 0 ; i < n ; i++){
            if(vis[i]) continue;
            if(b[i][j]) idx = i;
        }
        if(idx == -1) continue;
        vis[idx] = 1;
        h--;
        for(int i = 0 ; i < n ; i++){
            if(vis[i]) continue;
            if(b[i][j]) b[i] ^= b[idx];
        }
    }
    cout << 1LL * pow2[m - 1] * (pow2[n] - pow2[h] + mod) % mod << endl;
}

### Prompt

Your task is to create 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;
#define finish(x) return cout << x << endl, 0
#define ll long long

const int N = 301;

int n, m, pow2[N], mod = 998244353;
vector <bitset <N> > b;

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    pow2[0] = 1;
    for(int i = 1 ; i < N ; i++){
        pow2[i] = 2 * pow2[i - 1] % mod;
    }
    cin >> n >> m;
    b.resize(n);
    for(int i = 0 ; i < n ; i++){
        for(int j = 0 ; j < m ; j++){
            int x;
            cin >> x;
            if(x) b[i].set(j);
        }
    }
    vector <int> vis(n);
    int h = n;
    for(int j = 0 ; j < m ; j++){
        int idx = -1;
        for(int i = 0 ; i < n ; i++){
            if(vis[i]) continue;
            if(b[i][j]) idx = i;
        }
        if(idx == -1) continue;
        vis[idx] = 1;
        h--;
        for(int i = 0 ; i < n ; i++){
            if(vis[i]) continue;
            if(b[i][j]) b[i] ^= b[idx];
        }
    }
    cout << 1LL * pow2[m - 1] * (pow2[n] - pow2[h] + mod) % mod << endl;
}

```