#include<bits/stdc++.h>
using namespace std;
const int p=998244353;
int a[305][305],b[305][305];
int read() {
	char c=getchar(); int x=0,f=1;
	while (c<'0' || c>'9') {
		if (c=='-') f=-1;
		c=getchar();
	}
	while (c>='0' && c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
int fpow(int x,int k) {
	int ans=1;
	while (k) {
		if (k&1) ans=1LL*ans*x%p;
		x=1LL*x*x%p;
		k>>=1;
	}
	return ans;
}
int main() {
	int n=read(),m=read();
	for (int i=1; i<=n; i++) {
		for (int j=1; j<=m; j++) {
			a[i][j]=read();
		}
	}
	memset(b,0,sizeof(b));
	for (int i=1; i<=n; i++) {
		for (int j=m; j>=1; j--) {
			if (a[i][j]) {
				if (b[j][j]) for (int k=1; k<=m; k++) a[i][k]^=b[j][k];
				else {
					for (int k=1; k<=m; k++) b[j][k]=a[i][k];
					break;
				}
			}
		}
	}
	int cnt=0;
	for (int j=1; j<=m; j++) if (b[j][j]) ++cnt;
	printf("%d\n", ((fpow(2,n+m-1)-fpow(2,n+m-cnt-1))%p+p)%p);
	return 0;
}
### Prompt

Generate 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;
const int p=998244353;
int a[305][305],b[305][305];
int read() {
	char c=getchar(); int x=0,f=1;
	while (c<'0' || c>'9') {
		if (c=='-') f=-1;
		c=getchar();
	}
	while (c>='0' && c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
int fpow(int x,int k) {
	int ans=1;
	while (k) {
		if (k&1) ans=1LL*ans*x%p;
		x=1LL*x*x%p;
		k>>=1;
	}
	return ans;
}
int main() {
	int n=read(),m=read();
	for (int i=1; i<=n; i++) {
		for (int j=1; j<=m; j++) {
			a[i][j]=read();
		}
	}
	memset(b,0,sizeof(b));
	for (int i=1; i<=n; i++) {
		for (int j=m; j>=1; j--) {
			if (a[i][j]) {
				if (b[j][j]) for (int k=1; k<=m; k++) a[i][k]^=b[j][k];
				else {
					for (int k=1; k<=m; k++) b[j][k]=a[i][k];
					break;
				}
			}
		}
	}
	int cnt=0;
	for (int j=1; j<=m; j++) if (b[j][j]) ++cnt;
	printf("%d\n", ((fpow(2,n+m-1)-fpow(2,n+m-cnt-1))%p+p)%p);
	return 0;
}
```