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

const int N = 305, M = 998244353;

int n, m, c;
bitset<N> b[N];

int f(int x){
    int r = 1;
    for(; x--; ) r = r * 2 % M;
    return r;
}

int main(){
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i++){
        for(int j = 0, x; j < m; j++){
            scanf("%d", &x);
            b[i][j] = x;
        }
    }
    for(int i = 0; i < m; i++){
        int k = -1;
        for(int j = c; j < n; j++) if(b[j][i]){ k = j; break; }
        if(k < 0) continue;
        swap(b[c], b[k]);
        for(int j = c + 1; j < n; j++) if(b[j][i]) b[j] ^= b[c];
        c++;
    }
    printf("%lld\n", 1LL * f(n + m - 1 - c) * (f(c) + M - 1) % M);
}
### Prompt

Create a solution in cpp for 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;

const int N = 305, M = 998244353;

int n, m, c;
bitset<N> b[N];

int f(int x){
    int r = 1;
    for(; x--; ) r = r * 2 % M;
    return r;
}

int main(){
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i++){
        for(int j = 0, x; j < m; j++){
            scanf("%d", &x);
            b[i][j] = x;
        }
    }
    for(int i = 0; i < m; i++){
        int k = -1;
        for(int j = c; j < n; j++) if(b[j][i]){ k = j; break; }
        if(k < 0) continue;
        swap(b[c], b[k]);
        for(int j = c + 1; j < n; j++) if(b[j][i]) b[j] ^= b[c];
        c++;
    }
    printf("%lld\n", 1LL * f(n + m - 1 - c) * (f(c) + M - 1) % M);
}
```