#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#define LL long long
using namespace std;
const LL mod=998244353;
LL n,m,a[310][310],cnt=0;
LL c[310][310];
LL Pow(LL a,LL b)
{
	LL ans=1;
	while(b)
	{
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%lld %lld",&n,&m);
	for(LL i=1;i<=n;i++)
	{
		for(LL j=1;j<=m;j++) scanf("%lld",&a[i][j]);
		for(LL j=1;j<=m;j++)
		{
			if(!a[i][j]) continue;
			if(!c[j][j])
			{
				for(LL k=j;k<=m;k++) c[j][k]=a[i][k];
				cnt++;break;
			}
			else for(LL k=j;k<=m;k++) a[i][k]^=c[j][k];
		}
	}
	printf("%lld",(Pow(2,n)-Pow(2,n-cnt)+mod)%mod*Pow(2,m-1)%mod);
}
### Prompt

Your task is to create a CPP solution 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<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#define LL long long
using namespace std;
const LL mod=998244353;
LL n,m,a[310][310],cnt=0;
LL c[310][310];
LL Pow(LL a,LL b)
{
	LL ans=1;
	while(b)
	{
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%lld %lld",&n,&m);
	for(LL i=1;i<=n;i++)
	{
		for(LL j=1;j<=m;j++) scanf("%lld",&a[i][j]);
		for(LL j=1;j<=m;j++)
		{
			if(!a[i][j]) continue;
			if(!c[j][j])
			{
				for(LL k=j;k<=m;k++) c[j][k]=a[i][k];
				cnt++;break;
			}
			else for(LL k=j;k<=m;k++) a[i][k]^=c[j][k];
		}
	}
	printf("%lld",(Pow(2,n)-Pow(2,n-cnt)+mod)%mod*Pow(2,m-1)%mod);
}
```