#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 998244353
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
int n,m,a[305][305];


int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin>>a[i][j];
	int nxt = 1;
	for(int j=1;j<=m;j++){
		int pck = -1;
		for(int x=nxt;x<=n;x++){
			if(a[x][j] == 1){
				pck = x; break;
			}
		}
		if(pck == -1) continue;
		repn(j,m) swap(a[nxt][j],a[pck][j]);
		for(int x=nxt+1;x<=n;x++){
			if(a[x][j] == 1){
				for(int jj=j;jj<=m;jj++){
					a[x][jj] ^= a[nxt][jj];
				}
			}
		}
		nxt++;
	}
	int ans = n+1-nxt; 
	ll A = 1;
	rep(i,ans) A = A*2LL%mod;
	ll B = 1;
	rep(i,n) B = B*2LL%mod;
	B -= A;
	rep(i,m-1) B = B*2LL%mod;
	cout << (B%mod+mod)%mod << endl;
}
### Prompt

Create a solution in cpp for 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;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 998244353
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
int n,m,a[305][305];


int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin>>a[i][j];
	int nxt = 1;
	for(int j=1;j<=m;j++){
		int pck = -1;
		for(int x=nxt;x<=n;x++){
			if(a[x][j] == 1){
				pck = x; break;
			}
		}
		if(pck == -1) continue;
		repn(j,m) swap(a[nxt][j],a[pck][j]);
		for(int x=nxt+1;x<=n;x++){
			if(a[x][j] == 1){
				for(int jj=j;jj<=m;jj++){
					a[x][jj] ^= a[nxt][jj];
				}
			}
		}
		nxt++;
	}
	int ans = n+1-nxt; 
	ll A = 1;
	rep(i,ans) A = A*2LL%mod;
	ll B = 1;
	rep(i,n) B = B*2LL%mod;
	B -= A;
	rep(i,m-1) B = B*2LL%mod;
	cout << (B%mod+mod)%mod << endl;
}
```