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

#define rep(i, N) for (int i = 0; i < (N); i++)
#define all(a) (a).begin(), (a).end()
#define pb push_back

using ll = long long;
using i_i = tuple<int, int>;

const int MOD = 998244353;

int main() {
    int N, M;
    cin >> N >> M;
    vector<vector<int>> a(N, vector<int>(M));
    rep(i, N) rep(j, M) scanf("%d", &a[i][j]);
    int k = 0;
    rep(j, M) {
        int i0;
        for (i0 = k; i0 < N && !a[i0][j]; i0++);
        if (i0 == N) continue;
        swap(a[k], a[i0]);
        for (int i = k + 1; i < N; i++)
            if (a[i][j])
                rep(j, M)
                    a[i][j] ^= a[k][j];
        k++;
    }
    int rank = k;
    vector<ll> pow2(N + M + 1);
    pow2[0] = 1;
    rep(x, N + M) pow2[x + 1] = pow2[x] * 2 % MOD;
    ll ans = (pow2[N] - pow2[N - rank]) * pow2[M - 1] % MOD;
    cout << (ans + MOD) % MOD << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 rep(i, N) for (int i = 0; i < (N); i++)
#define all(a) (a).begin(), (a).end()
#define pb push_back

using ll = long long;
using i_i = tuple<int, int>;

const int MOD = 998244353;

int main() {
    int N, M;
    cin >> N >> M;
    vector<vector<int>> a(N, vector<int>(M));
    rep(i, N) rep(j, M) scanf("%d", &a[i][j]);
    int k = 0;
    rep(j, M) {
        int i0;
        for (i0 = k; i0 < N && !a[i0][j]; i0++);
        if (i0 == N) continue;
        swap(a[k], a[i0]);
        for (int i = k + 1; i < N; i++)
            if (a[i][j])
                rep(j, M)
                    a[i][j] ^= a[k][j];
        k++;
    }
    int rank = k;
    vector<ll> pow2(N + M + 1);
    pow2[0] = 1;
    rep(x, N + M) pow2[x + 1] = pow2[x] * 2 % MOD;
    ll ans = (pow2[N] - pow2[N - rank]) * pow2[M - 1] % MOD;
    cout << (ans + MOD) % MOD << endl;
}

```