#include<iostream>
#include<bitset>
using namespace std;

const int mod=998244353;

bitset<300> d[300],a[300];

int modpow(int x,int p){
	int res=1;
	while(p){
		if(p&1) res=1L*res*x%mod;
		x=1L*x*x%mod;
		p>>=1;
	}
	return res;
}

int main(){
	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[j][i]=x;
		}
	}
	
	int r=0;
	for(int i=0;i<m;i++){
		for(int j=n-1;j>=0;j--){
			if(a[i][j]==1){
				if(!d[j][j]) d[j]=a[i],r++;
				a[i]^=d[j];
			}
		}
	}
	cout << 1L*modpow(2,n-1)*(modpow(2,m)-modpow(2,m-r)+mod)%mod << endl;
}

### 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<iostream>
#include<bitset>
using namespace std;

const int mod=998244353;

bitset<300> d[300],a[300];

int modpow(int x,int p){
	int res=1;
	while(p){
		if(p&1) res=1L*res*x%mod;
		x=1L*x*x%mod;
		p>>=1;
	}
	return res;
}

int main(){
	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[j][i]=x;
		}
	}
	
	int r=0;
	for(int i=0;i<m;i++){
		for(int j=n-1;j>=0;j--){
			if(a[i][j]==1){
				if(!d[j][j]) d[j]=a[i],r++;
				a[i]^=d[j];
			}
		}
	}
	cout << 1L*modpow(2,n-1)*(modpow(2,m)-modpow(2,m-r)+mod)%mod << endl;
}

```