/* cerberus97 - Hanit Banga */

#include <iostream>
#include <iomanip>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <bitset>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 310, mod = 998244353;

bitset<N> a[N];
ll p2[N];

int rref(int n, int m);

int main() {
	fast_cin();
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < m; ++j) {
			int temp; cin >> temp;
			a[i][j] = temp;
		}
	}
	p2[0] = 1;
	for (int i = 1; i < N; ++i) {
		p2[i] = (p2[i - 1] * 2) % mod;
	}
	int fvar = rref(n, m);
	ll ans = p2[n - 1] * (p2[m] - p2[fvar] + mod);
	cout << ans % mod << endl;
}

int rref(int n, int m) {
	int sr = 0, pivots = 0;
	for (int j = 0; j < m; ++j) {
		for (int i = sr; i < n; ++i) {
			if (a[i][j]) {
				swap(a[i], a[sr]);
				break;
			}
		}
		if (!a[sr][j]) {
			continue;
		}
		++pivots;
		for (int i = sr + 1; i < n; ++i) {
			if (a[i][j]) {
				a[i] ^= a[sr];
			}
		}
		++sr;
	}
	return m - pivots;
}
### Prompt

Please provide a Cpp coded solution 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
/* cerberus97 - Hanit Banga */

#include <iostream>
#include <iomanip>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <bitset>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 310, mod = 998244353;

bitset<N> a[N];
ll p2[N];

int rref(int n, int m);

int main() {
	fast_cin();
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < m; ++j) {
			int temp; cin >> temp;
			a[i][j] = temp;
		}
	}
	p2[0] = 1;
	for (int i = 1; i < N; ++i) {
		p2[i] = (p2[i - 1] * 2) % mod;
	}
	int fvar = rref(n, m);
	ll ans = p2[n - 1] * (p2[m] - p2[fvar] + mod);
	cout << ans % mod << endl;
}

int rref(int n, int m) {
	int sr = 0, pivots = 0;
	for (int j = 0; j < m; ++j) {
		for (int i = sr; i < n; ++i) {
			if (a[i][j]) {
				swap(a[i], a[sr]);
				break;
			}
		}
		if (!a[sr][j]) {
			continue;
		}
		++pivots;
		for (int i = sr + 1; i < n; ++i) {
			if (a[i][j]) {
				a[i] ^= a[sr];
			}
		}
		++sr;
	}
	return m - pivots;
}
```