#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define sqr(x) ((x)*(x))
#define mp make_pair
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline ll read(){
	ll x = 0; char ch = gc(); bool positive = 1;
	for (; !isdigit(ch); ch = gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = gc())	x = x * 10 + ch - '0';
	return positive ? x : -x;
}
inline void write(ll a){
    if(a<0){
    	a=-a; putchar('-');
	}
    if(a>=10)write(a/10);
    putchar('0'+a%10);
}
inline void wri(ll a){write(a); putchar(' ');}
inline void writeln(ll a){write(a); puts("");}
const int N=300,mod=998244353;
int get(int n){
	int ans=1;
	for(int i=1;i<=n;i++)ans=ans*2%mod;
	return ans;
}
bitset<N> s[N];
signed main(){
	int n=read(),m=read();
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++)s[i][j]=read();
	}
	int rk=0;
	for(int i=0,j=0;i<m;i++){
		for(int k=j;k<n;k++)if(s[k][i]){swap(s[k],s[j]); break;}
		if(!s[j][i])continue; else rk++;
		for(int k=j+1;k<n;k++)if(s[k][i])s[k]^=s[j];
		j++;
	}
	int ans=get(n)-get(n-rk)+mod;
	cout<<(ll)ans*get(m-1)%mod<<endl;
}

### Prompt

Please create a solution in CPP 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 sqr(x) ((x)*(x))
#define mp make_pair
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline ll read(){
	ll x = 0; char ch = gc(); bool positive = 1;
	for (; !isdigit(ch); ch = gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = gc())	x = x * 10 + ch - '0';
	return positive ? x : -x;
}
inline void write(ll a){
    if(a<0){
    	a=-a; putchar('-');
	}
    if(a>=10)write(a/10);
    putchar('0'+a%10);
}
inline void wri(ll a){write(a); putchar(' ');}
inline void writeln(ll a){write(a); puts("");}
const int N=300,mod=998244353;
int get(int n){
	int ans=1;
	for(int i=1;i<=n;i++)ans=ans*2%mod;
	return ans;
}
bitset<N> s[N];
signed main(){
	int n=read(),m=read();
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++)s[i][j]=read();
	}
	int rk=0;
	for(int i=0,j=0;i<m;i++){
		for(int k=j;k<n;k++)if(s[k][i]){swap(s[k],s[j]); break;}
		if(!s[j][i])continue; else rk++;
		for(int k=j+1;k<n;k++)if(s[k][i])s[k]^=s[j];
		j++;
	}
	int ans=get(n)-get(n-rk)+mod;
	cout<<(ll)ans*get(m-1)%mod<<endl;
}

```