#include <cstdio>
#include <cstring>
#include <algorithm>
#include <bitset>
#define P 998244353

using namespace std;

typedef long long ll;

bitset <305> bt[305];

int n, m;

ll _2[777];

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++) {
            int v;
            scanf("%d", &v);
            if(v == 1) bt[i - 1][j - 1] = 1;
        }
    for(int i = 0; i < 777; i++) _2[i] = !i ? 1 : _2[i - 1] * 2 % P;
    int nw = 0;
    for(int i = 0; i < m; i++) {
        for(int j = nw; j < n; j++)
            if(bt[j][i]) {
                swap(bt[j], bt[nw]);
                break;
            }
        if(!bt[nw][i]) continue;
        for(int j = nw + 1; j < n; j++)
            if(bt[j][i]) bt[j] ^= bt[nw];
        nw++;
    }
    int _1 = nw, _0 = n - nw;
    ll ans = _2[m - 1] * (_2[_1] - 1) % P * _2[_0] % P;
    printf("%lld\n", ans);
    return 0;
}
### Prompt

Generate 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 <cstdio>
#include <cstring>
#include <algorithm>
#include <bitset>
#define P 998244353

using namespace std;

typedef long long ll;

bitset <305> bt[305];

int n, m;

ll _2[777];

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++) {
            int v;
            scanf("%d", &v);
            if(v == 1) bt[i - 1][j - 1] = 1;
        }
    for(int i = 0; i < 777; i++) _2[i] = !i ? 1 : _2[i - 1] * 2 % P;
    int nw = 0;
    for(int i = 0; i < m; i++) {
        for(int j = nw; j < n; j++)
            if(bt[j][i]) {
                swap(bt[j], bt[nw]);
                break;
            }
        if(!bt[nw][i]) continue;
        for(int j = nw + 1; j < n; j++)
            if(bt[j][i]) bt[j] ^= bt[nw];
        nw++;
    }
    int _1 = nw, _0 = n - nw;
    ll ans = _2[m - 1] * (_2[_1] - 1) % P * _2[_0] % P;
    printf("%lld\n", ans);
    return 0;
}
```