#include<bits/stdc++.h>
using namespace std;
const int MAXN=310;
const int mod=998244353;
bitset<MAXN>a[MAXN],b[MAXN];
int n,m,x,ans;
long long poww(long long x,long long y)
{
	long long ans=1;
	for (;y;y>>=1,x=(x*x)%mod) if (y&1) ans=(ans*x)%mod;
	return ans;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
	{
	for (int j=1;j<=m;j++) 
	{
		scanf("%d",&x);
		if (x==1) a[i].set(j);
	}
	}    for (int i=1;i<=n;i++)
    {
    	for (int j=1;j<=m;j++) 
    		if (!a[i][j]) continue;
    	else
    		if (!b[j].any()) {
    			b[j]=a[i]; break;
    		}
    		else a[i]^=b[j];
    }
    for (int i=1;i<=max(n,m);i++) if (b[i].any()) ans++;
    cout << (poww(2,n)-poww(2,n-ans)+mod)%mod*poww(2,m-1)%mod << endl; 
}
### Prompt

In CPP, your task is to solve 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;
const int MAXN=310;
const int mod=998244353;
bitset<MAXN>a[MAXN],b[MAXN];
int n,m,x,ans;
long long poww(long long x,long long y)
{
	long long ans=1;
	for (;y;y>>=1,x=(x*x)%mod) if (y&1) ans=(ans*x)%mod;
	return ans;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
	{
	for (int j=1;j<=m;j++) 
	{
		scanf("%d",&x);
		if (x==1) a[i].set(j);
	}
	}    for (int i=1;i<=n;i++)
    {
    	for (int j=1;j<=m;j++) 
    		if (!a[i][j]) continue;
    	else
    		if (!b[j].any()) {
    			b[j]=a[i]; break;
    		}
    		else a[i]^=b[j];
    }
    for (int i=1;i<=max(n,m);i++) if (b[i].any()) ans++;
    cout << (poww(2,n)-poww(2,n-ans)+mod)%mod*poww(2,m-1)%mod << endl; 
}
```