#include<bits/stdc++.h>
using namespace std;
#define ll long long
int rd(){
    int a=0;char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))a=(a<<1)+(a<<3)+(ch^48),ch=getchar();
    return a;
}
ll rdll(){
    ll a=0;char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))a=(a<<1)+(a<<3)+(ch^48),ch=getchar();
    return a;
}

const int N=304;
const ll p=998244353;
int n,m,x;
bitset<N>a[N],d[N];
ll expo(ll a,ll b){
    ll c=1;
    while(b){
        if(b&1)c=c*a%p;
        b>>=1,a=a*a%p;
    }
    return c;
}
void dbg(const bitset<N> &a){
	for(int i=1;i<=m;i++)printf(" %d",a[i]);printf("\n");
}
int main(){
	n=rd(),m=rd();
	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)a[i][j]=rd();
	for(int i=1;i<=n;i++)for(int j=m;j;j--)if(a[i][j]){//printf("a[%d]:",i);dbg(a[i]);printf("d[%d]:",j);dbg(d[j]);
	    if(d[j][j])a[i]^=d[j];
	    else{d[j]=a[i];break;}
	}
	for(int i=1;i<=m;i++)x+=d[i][i];//printf("x=%d\n",x);
	printf("%lld",((expo(2,m)-expo(2,m-x)+p)%p)*expo(2,n-1)%p);
	return 0;
}
### 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;
#define ll long long
int rd(){
    int a=0;char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))a=(a<<1)+(a<<3)+(ch^48),ch=getchar();
    return a;
}
ll rdll(){
    ll a=0;char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))a=(a<<1)+(a<<3)+(ch^48),ch=getchar();
    return a;
}

const int N=304;
const ll p=998244353;
int n,m,x;
bitset<N>a[N],d[N];
ll expo(ll a,ll b){
    ll c=1;
    while(b){
        if(b&1)c=c*a%p;
        b>>=1,a=a*a%p;
    }
    return c;
}
void dbg(const bitset<N> &a){
	for(int i=1;i<=m;i++)printf(" %d",a[i]);printf("\n");
}
int main(){
	n=rd(),m=rd();
	for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)a[i][j]=rd();
	for(int i=1;i<=n;i++)for(int j=m;j;j--)if(a[i][j]){//printf("a[%d]:",i);dbg(a[i]);printf("d[%d]:",j);dbg(d[j]);
	    if(d[j][j])a[i]^=d[j];
	    else{d[j]=a[i];break;}
	}
	for(int i=1;i<=m;i++)x+=d[i][i];//printf("x=%d\n",x);
	printf("%lld",((expo(2,m)-expo(2,m-x)+p)%p)*expo(2,n-1)%p);
	return 0;
}
```