/*


*/
#include <cstdio>
#include <cctype>
#include <bitset>
#include <algorithm>
#define gc() getchar()
#define mod 998244353
typedef long long LL;
const int N=305;

std::bitset<N> A[N],B[N];

inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-48,c=gc());
	return now;
}
inline int FP(int x,int k)
{
	int t=1;
	for(; k; k>>=1,x=1ll*x*x%mod)
		if(k&1) t=1ll*t*x%mod;
	return t;
}

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()==1;
	int r=0;
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j)
			if(A[i][j])//x>>j&1
			{
				if(B[j][j]) A[i]^=B[j];
				else
				{
					B[j]=A[i];
					++r; break;
				}
			}
	printf("%d\n",(FP(2,n+m-1)+mod-FP(2,n-r+m-1))%mod);

	return 0;
}
### Prompt

Please create a solution in CPP 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 <cctype>
#include <bitset>
#include <algorithm>
#define gc() getchar()
#define mod 998244353
typedef long long LL;
const int N=305;

std::bitset<N> A[N],B[N];

inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-48,c=gc());
	return now;
}
inline int FP(int x,int k)
{
	int t=1;
	for(; k; k>>=1,x=1ll*x*x%mod)
		if(k&1) t=1ll*t*x%mod;
	return t;
}

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()==1;
	int r=0;
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j)
			if(A[i][j])//x>>j&1
			{
				if(B[j][j]) A[i]^=B[j];
				else
				{
					B[j]=A[i];
					++r; break;
				}
			}
	printf("%d\n",(FP(2,n+m-1)+mod-FP(2,n-r+m-1))%mod);

	return 0;
}
```