#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int64_t i=0;i < (int64_t)(n);i++)

#define FILE_IN "post.in"
#define FILE_OUT "post.out"
#define ofile freopen(FILE_IN,"r",stdin);freopen(FILE_OUT,"w",stdout)
#define fio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define nfio cin.tie(0);cout.tie(0)
#define max(x,y) (((x)>(y))?(x):(y))
#define min(x,y) (((x)<(y))?(x):(y))
#define ord(a,b,c) ((a>=b)and(b>=c))
#define MOD (ll(998244353))
#define MAX 300001
#define mag 320
#define p1 first
#define p2 second.first
#define p3 second.second
#define pow2(x) (ll(1)<<x)
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pii;
typedef pair<ll,pii> piii;
#define endl "\n"
#define finish(s) {cout<<s;return 0;}


ll n,m,a[310][310];
ll od[310],ev[310],bit[310],xxj[310];
int main()
{
	fio;
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			cin>>a[i][j],a[i][j]&=1;
	if(n<m)
	{
		for(int i=1;i<=n;i++)
			for(int j=i+1;j<=m;j++)
				swap(a[i][j],a[j][i]);
		swap(n,m);
	}
 
	int u=max(n,m);
	od[1]=1;ev[1]=3;
	for(int i=2;i<=u;i++)
	{
		od[i]=(ev[i-1]+3ll*od[i-1]%MOD)%MOD;
		ev[i]=(od[i-1]+3ll*ev[i-1]%MOD)%MOD;
	}
	ll cnt=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
			if(a[i][j])
			{
				if(xxj[j])
					for(int k=j;k<=m;k++)
						a[i][k]^=a[xxj[j]][k];
				else {xxj[j]=i;cnt++;break;}
			}
	}
	bit[0]=1;
	for(int i=1;i<=u;i++) bit[i]=2ll*bit[i-1]%MOD;
	int bs=1ll*bit[n-cnt]*bit[m-cnt]%MOD;
	cout<<1ll*od[cnt]*bs%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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int64_t i=0;i < (int64_t)(n);i++)

#define FILE_IN "post.in"
#define FILE_OUT "post.out"
#define ofile freopen(FILE_IN,"r",stdin);freopen(FILE_OUT,"w",stdout)
#define fio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define nfio cin.tie(0);cout.tie(0)
#define max(x,y) (((x)>(y))?(x):(y))
#define min(x,y) (((x)<(y))?(x):(y))
#define ord(a,b,c) ((a>=b)and(b>=c))
#define MOD (ll(998244353))
#define MAX 300001
#define mag 320
#define p1 first
#define p2 second.first
#define p3 second.second
#define pow2(x) (ll(1)<<x)
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pii;
typedef pair<ll,pii> piii;
#define endl "\n"
#define finish(s) {cout<<s;return 0;}


ll n,m,a[310][310];
ll od[310],ev[310],bit[310],xxj[310];
int main()
{
	fio;
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			cin>>a[i][j],a[i][j]&=1;
	if(n<m)
	{
		for(int i=1;i<=n;i++)
			for(int j=i+1;j<=m;j++)
				swap(a[i][j],a[j][i]);
		swap(n,m);
	}
 
	int u=max(n,m);
	od[1]=1;ev[1]=3;
	for(int i=2;i<=u;i++)
	{
		od[i]=(ev[i-1]+3ll*od[i-1]%MOD)%MOD;
		ev[i]=(od[i-1]+3ll*ev[i-1]%MOD)%MOD;
	}
	ll cnt=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
			if(a[i][j])
			{
				if(xxj[j])
					for(int k=j;k<=m;k++)
						a[i][k]^=a[xxj[j]][k];
				else {xxj[j]=i;cnt++;break;}
			}
	}
	bit[0]=1;
	for(int i=1;i<=u;i++) bit[i]=2ll*bit[i-1]%MOD;
	int bs=1ll*bit[n-cnt]*bit[m-cnt]%MOD;
	cout<<1ll*od[cnt]*bs%MOD;
}
```