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

vector<int> operator^(const vector<int> &a, const vector<int> &b) {
    vector<int> res((int)a.size());
    for (int i = 0; i < (int)a.size(); ++i)
        res[i] = (a[i] ^ b[i]);
    return res;
}

bool operator<(const vector<int> &a, const vector<int> &b) {
    for (int i = 0; i < (int)a.size(); ++i) {
        if (a[i] == b[i]) continue;
        return a[i] < b[i];
    }
    return false;
}

const int mod = 998244353;

int main() {
    int n, m; scanf("%d%d", &n, &m);
    vector<vector<int>> a(n, vector<int>(m));
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j)
            scanf("%d", &a[i][j]);
    }
    vector<vector<int>> basis;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < (int)basis.size(); ++j)
            a[i] = min(a[i], a[i] ^ basis[j]);
        if (count(a[i].begin(), a[i].end(), 1) > 0)
            basis.push_back(a[i]);
    }
    int way = 1;
    for (int i = 0; i < n - (int)basis.size(); ++i)
        way = way * 2 % mod;
    int ans = 1;
    for (int i = 0; i < n; ++i)
        ans = ans * 2 % mod;
    ans = (ans + mod - way) % mod;
    int mult = 1;
    for (int i = 0; i < m - 1; ++i) 
        mult = mult * 2 % mod;
    printf("%d\n", ans * 1ll * mult % mod);
}

### 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;

vector<int> operator^(const vector<int> &a, const vector<int> &b) {
    vector<int> res((int)a.size());
    for (int i = 0; i < (int)a.size(); ++i)
        res[i] = (a[i] ^ b[i]);
    return res;
}

bool operator<(const vector<int> &a, const vector<int> &b) {
    for (int i = 0; i < (int)a.size(); ++i) {
        if (a[i] == b[i]) continue;
        return a[i] < b[i];
    }
    return false;
}

const int mod = 998244353;

int main() {
    int n, m; scanf("%d%d", &n, &m);
    vector<vector<int>> a(n, vector<int>(m));
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j)
            scanf("%d", &a[i][j]);
    }
    vector<vector<int>> basis;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < (int)basis.size(); ++j)
            a[i] = min(a[i], a[i] ^ basis[j]);
        if (count(a[i].begin(), a[i].end(), 1) > 0)
            basis.push_back(a[i]);
    }
    int way = 1;
    for (int i = 0; i < n - (int)basis.size(); ++i)
        way = way * 2 % mod;
    int ans = 1;
    for (int i = 0; i < n; ++i)
        ans = ans * 2 % mod;
    ans = (ans + mod - way) % mod;
    int mult = 1;
    for (int i = 0; i < m - 1; ++i) 
        mult = mult * 2 % mod;
    printf("%d\n", ans * 1ll * mult % mod);
}

```