#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define MOD 998244353

const int N = 300;
int n, m;
int A[N + 7][N + 7];

ll power(ll a, int k){
	ll ans = 1;
	while(k > 0){
		if(k&1)
			ans = ans * a % MOD;
		a = a * a % MOD;
		k >>= 1;
	}
	return ans;
}

int main(){
	cin >> n >> m;
	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++)
			cin >> A[i][j];

	int myi = 0;
	for(int j = 0; j < m && myi < n; j++){
		for(int i = myi; i < n; i++)
			if(A[i][j]){
				swap(A[i], A[myi]);
				break;
			}
		if(!A[myi][j])
			continue;
		for(int i = myi; i < n; i++)
			for(int jj = j; jj < m; jj++)
				A[i][jj] = A[i][jj] ^ (A[myi][jj] * A[i][j]);
		myi++; 
	}

	ll ans = power(2, n + m - 1);
	ll ans2 = power(2, n + m - 1 - myi);
	ans -= ans2;
	if(ans < 0)
		ans += MOD;
	cout << ans << "\n";
}
### 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 int
#define MOD 998244353

const int N = 300;
int n, m;
int A[N + 7][N + 7];

ll power(ll a, int k){
	ll ans = 1;
	while(k > 0){
		if(k&1)
			ans = ans * a % MOD;
		a = a * a % MOD;
		k >>= 1;
	}
	return ans;
}

int main(){
	cin >> n >> m;
	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++)
			cin >> A[i][j];

	int myi = 0;
	for(int j = 0; j < m && myi < n; j++){
		for(int i = myi; i < n; i++)
			if(A[i][j]){
				swap(A[i], A[myi]);
				break;
			}
		if(!A[myi][j])
			continue;
		for(int i = myi; i < n; i++)
			for(int jj = j; jj < m; jj++)
				A[i][jj] = A[i][jj] ^ (A[myi][jj] * A[i][j]);
		myi++; 
	}

	ll ans = power(2, n + m - 1);
	ll ans2 = power(2, n + m - 1 - myi);
	ans -= ans2;
	if(ans < 0)
		ans += MOD;
	cout << ans << "\n";
}
```