#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

const int MOD = 998244353;

const int MX = 307;

using bs = bitset<MX>;

bs a[MX];
int pw[2 * MX];
int main() {
#ifdef FASTIO
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
    int n, m;
    cin >> n >> m;
    pw[0] = 1;
    for (int i = 1; i < 2 * MX; i++) {
        pw[i] = (pw[i - 1] + pw[i - 1]) % MOD;
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            int x;
            cin >> x;
            if (x == 1) {
                a[i][j] = true;
            }
        }
    }

    int basis = 0;
    int ptr = 1;
    for (int i = 1; i <= m; i++) {
        int pos = ptr;
        while (pos <= n && a[pos][i] == false) {
            pos++;
        }
        if (pos == n + 1) {
            continue;
        }
        basis++;
        swap(a[pos], a[ptr]);
        for (int j = ptr + 1; j <= n; j++) {
            if (a[j][i]) {
                a[j] ^= a[ptr];
            }
        }
        ptr++;
    }

    int nz = (pw[n] - pw[n - basis] + MOD) % MOD;
    int ans = (1ll * nz * pw[m - 1]) % MOD;
    cout << ans << "\n";
    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
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

const int MOD = 998244353;

const int MX = 307;

using bs = bitset<MX>;

bs a[MX];
int pw[2 * MX];
int main() {
#ifdef FASTIO
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
    int n, m;
    cin >> n >> m;
    pw[0] = 1;
    for (int i = 1; i < 2 * MX; i++) {
        pw[i] = (pw[i - 1] + pw[i - 1]) % MOD;
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            int x;
            cin >> x;
            if (x == 1) {
                a[i][j] = true;
            }
        }
    }

    int basis = 0;
    int ptr = 1;
    for (int i = 1; i <= m; i++) {
        int pos = ptr;
        while (pos <= n && a[pos][i] == false) {
            pos++;
        }
        if (pos == n + 1) {
            continue;
        }
        basis++;
        swap(a[pos], a[ptr]);
        for (int j = ptr + 1; j <= n; j++) {
            if (a[j][i]) {
                a[j] ^= a[ptr];
            }
        }
        ptr++;
    }

    int nz = (pw[n] - pw[n - basis] + MOD) % MOD;
    int ans = (1ll * nz * pw[m - 1]) % MOD;
    cout << ans << "\n";
    return 0;
}


```