#include <bits/stdc++.h>
#define MOD 998244353
using namespace std;

#define rep(i, a, b) for(int i = (a); i < int(b); ++i)
#define rrep(i, a, b) for(int i = (a) - 1; i >= int(b); --i)
#define trav(it, v) for(auto &it : (v))
#define all(v) (v).begin(), (v).end()
#define what_is(x) cerr << #x << " is " << x << endl;
#define sz(x) (int)(x).size()

typedef double fl;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;

ll powerMod(ll a, ll x) {
    ll ans = 1;
    while(x--) {
        ans *= a;
        ans %= MOD;
    }
    return ans;
}

int main(){
	ios::sync_with_stdio(0);
    int N, M;
    cin >> N >> M;
    vector<vector<int>> a(N, vector<int>(M));
    rep(i,0,N)
        rep(j,0,M) {
            cin >> a[i][j];
        }
    int pos = 0;
    rep(j,0,M) {
        bool ok = 0;
        rep(i,pos,N) {
            if (a[i][j]) {
                swap(a[i], a[pos]);
                ok = 1;
                break;
            }
        }
        if (ok) {
            rep(i,pos+1,N) {
                if (!a[i][j])
                    continue;
                rep(j2,0,M) {
                    a[i][j2] ^= a[pos][j2];
                }
            }
            ++pos;
        }
    }
    int rank = pos;
    ll ans = powerMod(2, N+M-1);
    ans -= powerMod(2, N+M-1-rank);
    ans %= MOD;
    if (ans < 0)
        ans += MOD;
    cout << ans << endl;
}
### 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>
#define MOD 998244353
using namespace std;

#define rep(i, a, b) for(int i = (a); i < int(b); ++i)
#define rrep(i, a, b) for(int i = (a) - 1; i >= int(b); --i)
#define trav(it, v) for(auto &it : (v))
#define all(v) (v).begin(), (v).end()
#define what_is(x) cerr << #x << " is " << x << endl;
#define sz(x) (int)(x).size()

typedef double fl;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;

ll powerMod(ll a, ll x) {
    ll ans = 1;
    while(x--) {
        ans *= a;
        ans %= MOD;
    }
    return ans;
}

int main(){
	ios::sync_with_stdio(0);
    int N, M;
    cin >> N >> M;
    vector<vector<int>> a(N, vector<int>(M));
    rep(i,0,N)
        rep(j,0,M) {
            cin >> a[i][j];
        }
    int pos = 0;
    rep(j,0,M) {
        bool ok = 0;
        rep(i,pos,N) {
            if (a[i][j]) {
                swap(a[i], a[pos]);
                ok = 1;
                break;
            }
        }
        if (ok) {
            rep(i,pos+1,N) {
                if (!a[i][j])
                    continue;
                rep(j2,0,M) {
                    a[i][j2] ^= a[pos][j2];
                }
            }
            ++pos;
        }
    }
    int rank = pos;
    ll ans = powerMod(2, N+M-1);
    ans -= powerMod(2, N+M-1-rank);
    ans %= MOD;
    if (ans < 0)
        ans += MOD;
    cout << ans << endl;
}
```