#include<bits/stdc++.h>
using namespace std;
 
#define ll long long
#define pb push_back
 
const int maxn = 3e2 + 20;
const int mod = 998244353;

bitset<maxn> a[maxn];

int ind[maxn];

int bpw(int a , int b)
{
	if(!b)
		return 1;
	int x = bpw(a , b / 2);

	x = 1LL * x * x % mod;
	if(b&1)
		x = 1LL * x * a % mod;

	return x;
}

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

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

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

	memset(ind , -1 , sizeof ind);

	int res = 0;
	for(int i = 0; i < n; i++)
	{
		for(int j = m - 1; j >= 0; j--)
			if(a[i][j] && ind[j] != -1)
				a[i] ^= a[ind[j]];

		int last = -1;
		for(int j = 0; j < m; j++)
			if(a[i][j])
				last = j;

		if(last == -1)
			continue;

		for(int j = m - 1; j >= 0; j--)
			if(ind[j] != -1 && a[ind[j]][last])
				a[ind[j]] ^= a[i];

		ind[last] = i;
		res++;
	}

	res = (1LL * bpw(2 , n - res) * (bpw(2 , res) - 1 + mod)) % mod;
	res = 1LL * res * bpw(2 , m - 1) % mod;

	cout << res << endl;
}

















### Prompt

Your task is to create 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;
 
#define ll long long
#define pb push_back
 
const int maxn = 3e2 + 20;
const int mod = 998244353;

bitset<maxn> a[maxn];

int ind[maxn];

int bpw(int a , int b)
{
	if(!b)
		return 1;
	int x = bpw(a , b / 2);

	x = 1LL * x * x % mod;
	if(b&1)
		x = 1LL * x * a % mod;

	return x;
}

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

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

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

	memset(ind , -1 , sizeof ind);

	int res = 0;
	for(int i = 0; i < n; i++)
	{
		for(int j = m - 1; j >= 0; j--)
			if(a[i][j] && ind[j] != -1)
				a[i] ^= a[ind[j]];

		int last = -1;
		for(int j = 0; j < m; j++)
			if(a[i][j])
				last = j;

		if(last == -1)
			continue;

		for(int j = m - 1; j >= 0; j--)
			if(ind[j] != -1 && a[ind[j]][last])
				a[ind[j]] ^= a[i];

		ind[last] = i;
		res++;
	}

	res = (1LL * bpw(2 , n - res) * (bpw(2 , res) - 1 + mod)) % mod;
	res = 1LL * res * bpw(2 , m - 1) % mod;

	cout << res << endl;
}

















```