#include<bits/stdc++.h>
#define Mod 998244353
#define N 305
using namespace std;
int n,m,x,ans,bin[N],vis[N];bitset<N>a[N],tmp[N];
int main(){
	bin[0]=1;
	for (int i=1;i<N;i++) bin[i]=1ll*bin[i-1]*2%Mod;
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			scanf("%d",&x);
			a[i][j]=x;
		}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			if (a[i][j]){
				if (vis[j])
					a[i]^=tmp[j];
				else {
					vis[j]=1;
					tmp[j]=a[i];
					break;
				}
			}
		}
	}
	for (int i=1;i<=m;i++) ans+=vis[i];
	int tmpp=(bin[n]-bin[n-ans]+Mod)%Mod;
	tmpp=1ll*tmpp*bin[m-1]%Mod;
	printf("%d\n",tmpp);
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#define Mod 998244353
#define N 305
using namespace std;
int n,m,x,ans,bin[N],vis[N];bitset<N>a[N],tmp[N];
int main(){
	bin[0]=1;
	for (int i=1;i<N;i++) bin[i]=1ll*bin[i-1]*2%Mod;
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			scanf("%d",&x);
			a[i][j]=x;
		}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			if (a[i][j]){
				if (vis[j])
					a[i]^=tmp[j];
				else {
					vis[j]=1;
					tmp[j]=a[i];
					break;
				}
			}
		}
	}
	for (int i=1;i<=m;i++) ans+=vis[i];
	int tmpp=(bin[n]-bin[n-ans]+Mod)%Mod;
	tmpp=1ll*tmpp*bin[m-1]%Mod;
	printf("%d\n",tmpp);
	return 0;
}
```