#include<bits/stdc++.h>
using namespace std;
const int P=998244353;
int n,m,ans,A[305][305],two[605];
void up(int &x){if (x>=P) x-=P;}

int xor_guass(int m, int n) //A是异或方程组系数矩阵 返回秩
{
    int i = 0, j = 0, k, r, u;
    while(i < m && j < n){//当前正在处理第i个方程,第j个变量
        r = i;
        for(int k = i; k < m; k++) if(A[k][j]){r = k; break;}
        if(A[r][j]){
             if(r != i) for(k = 0; k <= n; k++) swap(A[r][k], A[i][k]);
             //消元完成之后第i行的第一个非0列是第j列,且第u>i行的第j列全是0
            for(u = i + 1; u < m; u++) if(A[u][j])
                for(k = i; k <= n; k++) A[u][k] ^= A[i][k];
            i++;
        }
        j++;
    }
    return i;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++){
			int x;scanf("%d",&x);
			A[j-1][i-1]=x;
		}
	 
	two[0]=1;
	for (int i=1;i<=n+m;i++)
		two[i]=two[i-1]*2%P;
	ans=1ll*(two[n]-two[n-xor_guass(m,n)]+P)*two[m-1]%P; 
	printf("%d\n",ans);
}
### Prompt

Your challenge is to write 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 n,m,ans,A[305][305],two[605];
void up(int &x){if (x>=P) x-=P;}

int xor_guass(int m, int n) //A是异或方程组系数矩阵 返回秩
{
    int i = 0, j = 0, k, r, u;
    while(i < m && j < n){//当前正在处理第i个方程,第j个变量
        r = i;
        for(int k = i; k < m; k++) if(A[k][j]){r = k; break;}
        if(A[r][j]){
             if(r != i) for(k = 0; k <= n; k++) swap(A[r][k], A[i][k]);
             //消元完成之后第i行的第一个非0列是第j列,且第u>i行的第j列全是0
            for(u = i + 1; u < m; u++) if(A[u][j])
                for(k = i; k <= n; k++) A[u][k] ^= A[i][k];
            i++;
        }
        j++;
    }
    return i;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++){
			int x;scanf("%d",&x);
			A[j-1][i-1]=x;
		}
	 
	two[0]=1;
	for (int i=1;i<=n+m;i++)
		two[i]=two[i-1]*2%P;
	ans=1ll*(two[n]-two[n-xor_guass(m,n)]+P)*two[m-1]%P; 
	printf("%d\n",ans);
}
```