#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	int N,M; cin >> N >> M;
	const LL mod = 998244353;
	vector<bitset<300>> b(N,bitset<300>(0));
	repp(i,0,N) repp(j,0,M){
		int a; cin >> a;
		b[i][j] = a;
	}
	int k = 0;
	repp(j,0,M){
		int i = k;
		while(i < N && !b[i][j]) ++i;
		if(i == N) continue;
		swap(b[k],b[i]);
		repp(p,k+1,N) if(b[p][j]) b[p] ^= b[k];
		++k;
	}
	LL z = 1;
	repp(i,k,N) z = z*2%mod;
	LL ans = 1;
	repp(i,0,N) ans = ans*2%mod;
	ans = (ans+mod-z)%mod;
	repp(i,1,M) ans = ans*2%mod;
	cout << ans << 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;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	int N,M; cin >> N >> M;
	const LL mod = 998244353;
	vector<bitset<300>> b(N,bitset<300>(0));
	repp(i,0,N) repp(j,0,M){
		int a; cin >> a;
		b[i][j] = a;
	}
	int k = 0;
	repp(j,0,M){
		int i = k;
		while(i < N && !b[i][j]) ++i;
		if(i == N) continue;
		swap(b[k],b[i]);
		repp(p,k+1,N) if(b[p][j]) b[p] ^= b[k];
		++k;
	}
	LL z = 1;
	repp(i,k,N) z = z*2%mod;
	LL ans = 1;
	repp(i,0,N) ans = ans*2%mod;
	ans = (ans+mod-z)%mod;
	repp(i,1,M) ans = ans*2%mod;
	cout << ans << endl;
	return 0;
}

```