#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MD = 998244353;
using ll = long long;
int n, m, a, t;
bitset<305> bt[305];
ll e(ll n){
	ll res = 1;
	for(ll b = 2; n; n/=2,(b*=b)%=MD) if(n%2) (res *= b) %= MD;
	return res;
}
int main(){
	scanf("%d%d", &n, &m);
	rep(i,n) rep(j,m){
		scanf("%d", &a);
		bt[i][j] = a;
	}
	rep(i,m){
		for(int j = t+1; j < n; j++) if(bt[j][i]){
			if(bt[t][i]) bt[j] ^= bt[t];
			else swap(bt[t], bt[j]);
		}
		if(bt[t][i]) t++;
	}
	printf("%lld\n", (e(m) - e(m-t) + MD) * e(n-1) % MD);
}
### Prompt

Please formulate 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;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MD = 998244353;
using ll = long long;
int n, m, a, t;
bitset<305> bt[305];
ll e(ll n){
	ll res = 1;
	for(ll b = 2; n; n/=2,(b*=b)%=MD) if(n%2) (res *= b) %= MD;
	return res;
}
int main(){
	scanf("%d%d", &n, &m);
	rep(i,n) rep(j,m){
		scanf("%d", &a);
		bt[i][j] = a;
	}
	rep(i,m){
		for(int j = t+1; j < n; j++) if(bt[j][i]){
			if(bt[t][i]) bt[j] ^= bt[t];
			else swap(bt[t], bt[j]);
		}
		if(bt[t][i]) t++;
	}
	printf("%lld\n", (e(m) - e(m-t) + MD) * e(n-1) % MD);
}
```