#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)

using namespace std;

const int N=310,mod=998244353;
int n,m,r,sum,all,vis[N]; bitset<N> b[N];

int insert(bitset<N> x){
	rep (i,0,m-1)
		if (x.test(i))
			if (!vis[i]){vis[i]=1,b[i]=x; return 1;}
			else x^=b[i];
	return 0;
}

int main(){
	scanf("%d%d",&n,&m);
	rep (i,1,n){
		bitset<N> a;
		rep (j,1,m){
			int x; scanf("%d",&x);
			if (x) a.set(j-1);
		}
		r+=insert(a);
	}
	all=1;
	rep (i,1,n) all=(all+all)%mod;
	sum=1;
	rep (i,1,n-r) sum=(sum+sum)%mod;
	all=(all+mod-sum)%mod;
	rep (i,1,m-1) all=(all+all)%mod;
	printf("%d\n",all);
	return 0;
}
### 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>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)

using namespace std;

const int N=310,mod=998244353;
int n,m,r,sum,all,vis[N]; bitset<N> b[N];

int insert(bitset<N> x){
	rep (i,0,m-1)
		if (x.test(i))
			if (!vis[i]){vis[i]=1,b[i]=x; return 1;}
			else x^=b[i];
	return 0;
}

int main(){
	scanf("%d%d",&n,&m);
	rep (i,1,n){
		bitset<N> a;
		rep (j,1,m){
			int x; scanf("%d",&x);
			if (x) a.set(j-1);
		}
		r+=insert(a);
	}
	all=1;
	rep (i,1,n) all=(all+all)%mod;
	sum=1;
	rep (i,1,n-r) sum=(sum+sum)%mod;
	all=(all+mod-sum)%mod;
	rep (i,1,m-1) all=(all+all)%mod;
	printf("%d\n",all);
	return 0;
}
```