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

int n, m;
int a[305][305];
bool check[305];

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);

	cin >> n >> m;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) cin >> a[i][j];
	}
	
	// a[i][m + 1] = 0
	int rank = 0;
	for (int col = 1; col <= m; ++col) {
		int row = 0;
		for (int i = 1; i <= n; ++i) if (!check[i] && a[i][col]) row = i;
		if (!row) {
			++rank; continue;
		}
		check[row] = true;
		for (int i = 1; i <= n; ++i) if (i != row && a[i][col]) {
			for (int j = 1; j <= m; ++j) a[i][j] ^= a[row][j];
		}
	}
	
	// 2^(m + n - 1) - 2^(rank + n - 1)
	int tmp = 1, res = 1;
	static const int mod = 998244353;
	for (int i = 1; i <= m + n - 1; ++i) res = 2LL * res % mod;
	for (int i = 1; i <= rank + n - 1; ++i) tmp = 2LL * tmp % mod;
	
	res = (res - tmp) % mod;
	if (res < 0) res += mod;
	
	cout << res << endl;
}

### Prompt

Your challenge is to write 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
#include <bits/stdc++.h>
using namespace std;

int n, m;
int a[305][305];
bool check[305];

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);

	cin >> n >> m;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) cin >> a[i][j];
	}
	
	// a[i][m + 1] = 0
	int rank = 0;
	for (int col = 1; col <= m; ++col) {
		int row = 0;
		for (int i = 1; i <= n; ++i) if (!check[i] && a[i][col]) row = i;
		if (!row) {
			++rank; continue;
		}
		check[row] = true;
		for (int i = 1; i <= n; ++i) if (i != row && a[i][col]) {
			for (int j = 1; j <= m; ++j) a[i][j] ^= a[row][j];
		}
	}
	
	// 2^(m + n - 1) - 2^(rank + n - 1)
	int tmp = 1, res = 1;
	static const int mod = 998244353;
	for (int i = 1; i <= m + n - 1; ++i) res = 2LL * res % mod;
	for (int i = 1; i <= rank + n - 1; ++i) tmp = 2LL * tmp % mod;
	
	res = (res - tmp) % mod;
	if (res < 0) res += mod;
	
	cout << res << endl;
}

```