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

const int Mx = 300;
const int mod = 998244353;
ll p2[Mx + 1];

typedef bitset<Mx> * It;

ll f(It first, It last, const int pos) {

	if (pos < 0 || first == last) {
		assert(all_of(first, last, [](const bitset<Mx> &x) {
			return x == 0;
		}));
		return p2[last - first];
	}

	partition(first, last, [&pos](const bitset<Mx> &x) {
		return x[pos] == 0;
	});

	It pLast = prev(last);
	if ((*pLast)[pos] == 0)
		return f(first, last, pos - 1);
	for (It it = first; it < pLast; ++it)
		if ((*it)[pos])
			*it ^= *pLast;
	return f(first, pLast, pos - 1);
}

bitset<Mx> a[300];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);

	p2[0] = 1;
	for (int i = 1; i <= Mx; ++i)
		p2[i] = (p2[i - 1] * 2) % mod;

	int n, m;
	cin >> n >> m;

	for (int i = 0; i < n; ++i) {
		a[i] = 0;
		for (int j = 0; j < m; ++j) {
			int x;
			cin >> x;
			a[i][j] = x;
		}
	}

	ll res = f(a, a + n, m - 1);
	res = (p2[n] - res) * p2[m - 1];
	cout << (res % mod + mod) % mod;
	return 0;
}

### Prompt

Please create a solution in cpp 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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int Mx = 300;
const int mod = 998244353;
ll p2[Mx + 1];

typedef bitset<Mx> * It;

ll f(It first, It last, const int pos) {

	if (pos < 0 || first == last) {
		assert(all_of(first, last, [](const bitset<Mx> &x) {
			return x == 0;
		}));
		return p2[last - first];
	}

	partition(first, last, [&pos](const bitset<Mx> &x) {
		return x[pos] == 0;
	});

	It pLast = prev(last);
	if ((*pLast)[pos] == 0)
		return f(first, last, pos - 1);
	for (It it = first; it < pLast; ++it)
		if ((*it)[pos])
			*it ^= *pLast;
	return f(first, pLast, pos - 1);
}

bitset<Mx> a[300];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);

	p2[0] = 1;
	for (int i = 1; i <= Mx; ++i)
		p2[i] = (p2[i - 1] * 2) % mod;

	int n, m;
	cin >> n >> m;

	for (int i = 0; i < n; ++i) {
		a[i] = 0;
		for (int j = 0; j < m; ++j) {
			int x;
			cin >> x;
			a[i][j] = x;
		}
	}

	ll res = f(a, a + n, m - 1);
	res = (p2[n] - res) * p2[m - 1];
	cout << (res % mod + mod) % mod;
	return 0;
}

```