#include<cstdio>
#include<vector>
#include<set>

using namespace std;

const long long mod = 998244353;

vector<int> A[330];
int N, M;

void input(){
	scanf("%d%d", &N, &M);
	for(int i = 0; i < N; ++i){
		for(int j = 0; j < M; ++j){
			int x;
			scanf("%d", &x);
			A[i].push_back(x);
		}
	}
}

int get(vector<int> v){
	for(int i = 0; i < v.size(); ++i) if(v[i] != 0) return i;
	return -1;
}

vector<int> er(vector<int> v, vector<int> e){
	int i1 = get(v);
	int i2 = get(e);
	if(i1 != i2) return v;
	for(int i = 0; i < v.size(); ++i){
		v[i] ^= e[i];
	}
	return v;
}

set<vector<int> > se;

long long pw[330];

long long mul(long long a, long long b){
	return (a * b) % mod;
}

long long solve(){
	pw[0] = 1;
	for(int i = 1; i < 330; ++i){
		pw[i] = (pw[i - 1] * 2) % mod;
	}
	for(int i = 0; i < N; ++i){
		int id = get(A[i]);
		if(id == -1) continue;
		vector<int> v = A[i];
		for(set<vector<int> >::reverse_iterator it = se.rbegin(); it != se.rend(); ++it){
			v = er(v, *it);
		}
		if(get(v) != -1){
			se.insert(v);
		}
	}
	long long ans = pw[N - se.size()];
	ans = mul(ans, pw[se.size()] - 1);
	ans = mul(ans, pw[M - 1]);
	ans += mod;
	ans %= mod;
	return ans;
}

int main(){
	input();
	long long ans = solve();
	printf("%lld\n", ans);
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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<cstdio>
#include<vector>
#include<set>

using namespace std;

const long long mod = 998244353;

vector<int> A[330];
int N, M;

void input(){
	scanf("%d%d", &N, &M);
	for(int i = 0; i < N; ++i){
		for(int j = 0; j < M; ++j){
			int x;
			scanf("%d", &x);
			A[i].push_back(x);
		}
	}
}

int get(vector<int> v){
	for(int i = 0; i < v.size(); ++i) if(v[i] != 0) return i;
	return -1;
}

vector<int> er(vector<int> v, vector<int> e){
	int i1 = get(v);
	int i2 = get(e);
	if(i1 != i2) return v;
	for(int i = 0; i < v.size(); ++i){
		v[i] ^= e[i];
	}
	return v;
}

set<vector<int> > se;

long long pw[330];

long long mul(long long a, long long b){
	return (a * b) % mod;
}

long long solve(){
	pw[0] = 1;
	for(int i = 1; i < 330; ++i){
		pw[i] = (pw[i - 1] * 2) % mod;
	}
	for(int i = 0; i < N; ++i){
		int id = get(A[i]);
		if(id == -1) continue;
		vector<int> v = A[i];
		for(set<vector<int> >::reverse_iterator it = se.rbegin(); it != se.rend(); ++it){
			v = er(v, *it);
		}
		if(get(v) != -1){
			se.insert(v);
		}
	}
	long long ans = pw[N - se.size()];
	ans = mul(ans, pw[se.size()] - 1);
	ans = mul(ans, pw[M - 1]);
	ans += mod;
	ans %= mod;
	return ans;
}

int main(){
	input();
	long long ans = solve();
	printf("%lld\n", ans);
	return 0;
}

```