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

const int maxn = 310, P = 998244353;
int n, m;
bitset<maxn> a[maxn];
vector<bitset<maxn> > base;

int qp(int x, int y) {
    int z = 1;
    for (; y; y >>= 1, x = 1LL * x * x % P) {
        if (y & 1) z = 1LL * z * x % P;
    }
    return z;
}

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) {
        for (int j = m - 1, t; ~j; j--) {
            scanf("%d", &t);
            if (t) a[i].set(j);
        }
    }
    for (int i = 0; i < n; i++) {
        for (auto b : base) {
            auto c = b ^ a[i];
            for (int j = m - 1; ~j; j--) {
                if (c[j] < a[i][j]) { a[i] = c; break; }
                else if (c[j] > a[i][j]) break;
            }
        }
        if (a[i].any()) base.push_back(a[i]);
    }
    printf("%d\n", (qp(2, n + m - 1) - qp(2, n + m - 1 - base.size()) + P) % P);
    return 0;
}
### 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;

const int maxn = 310, P = 998244353;
int n, m;
bitset<maxn> a[maxn];
vector<bitset<maxn> > base;

int qp(int x, int y) {
    int z = 1;
    for (; y; y >>= 1, x = 1LL * x * x % P) {
        if (y & 1) z = 1LL * z * x % P;
    }
    return z;
}

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) {
        for (int j = m - 1, t; ~j; j--) {
            scanf("%d", &t);
            if (t) a[i].set(j);
        }
    }
    for (int i = 0; i < n; i++) {
        for (auto b : base) {
            auto c = b ^ a[i];
            for (int j = m - 1; ~j; j--) {
                if (c[j] < a[i][j]) { a[i] = c; break; }
                else if (c[j] > a[i][j]) break;
            }
        }
        if (a[i].any()) base.push_back(a[i]);
    }
    printf("%d\n", (qp(2, n + m - 1) - qp(2, n + m - 1 - base.size()) + P) % P);
    return 0;
}
```