#include <bits/stdc++.h>

using namespace std;

vector<int> a[333];
int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		a[i].resize(m);
		for (int j = 0; j < m; j++)
			scanf("%d", &a[i][j]);
	}
	int p = 998244353;
	int r = [](vector<int> *m, int R, int C) {
		int i = 0;
		for (int j = 0; j < C; j++) {
			int k;
			for (k = i; k < R; k++)
				if (m[k][j]) goto p;
			continue;
		p:
			swap(m[k], m[i]);
			for (k = 0; k < R; k++)
				if (k != i && m[k][j])
					for (int l = j; l < C; l++)
						m[k][l] ^= m[i][l];
			i++;
		}
		return i;
	}(a, n, m);
	int x = 0;
	for (int i = 0; i < 2; i++) {
		int y = 1;
		for (int j = 0; j < n + m - 1 - i * r; j++)
			(y += y) %= p;
		(x += i ? p - y : y) %= p;
	}
	cout << x << endl;
	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
#include <bits/stdc++.h>

using namespace std;

vector<int> a[333];
int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		a[i].resize(m);
		for (int j = 0; j < m; j++)
			scanf("%d", &a[i][j]);
	}
	int p = 998244353;
	int r = [](vector<int> *m, int R, int C) {
		int i = 0;
		for (int j = 0; j < C; j++) {
			int k;
			for (k = i; k < R; k++)
				if (m[k][j]) goto p;
			continue;
		p:
			swap(m[k], m[i]);
			for (k = 0; k < R; k++)
				if (k != i && m[k][j])
					for (int l = j; l < C; l++)
						m[k][l] ^= m[i][l];
			i++;
		}
		return i;
	}(a, n, m);
	int x = 0;
	for (int i = 0; i < 2; i++) {
		int y = 1;
		for (int j = 0; j < n + m - 1 - i * r; j++)
			(y += y) %= p;
		(x += i ? p - y : y) %= p;
	}
	cout << x << endl;
	return 0;
}

```