#include <bits/stdc++.h>
using namespace std;
const int maxn = 333, mod = 998244353;
bitset<maxn> e[maxn];
int power(int x){
	int v = 1;
	for(int i = 0; i < x; i++){
		v *= 2;
		if(v >= mod) v -= mod;
	}
	return v;
}
int main(){
	int n, m; cin >> n >> m; int rnk = 0;
	for(int i = 0; i < n; i++){
		bitset<maxn> now;
		for(int j = 0; j < m; j++){
			int v; cin >> v;
			now[j] = v == 1;
		}
		for(int j = 0; j < m; j++){
			if(e[j][j] && now[j]){
				now ^= e[j];
			} else if(now[j]){
				e[j] = now;
				rnk++;
				break;
			}
		}
	}
	int ans = power(n + m - 1);
	ans += mod - power(n + m - 1 - rnk);
	if(ans >= mod) ans -= mod;
	cout << ans << endl;
	return 0;
}


### Prompt

Please provide a Cpp coded solution to the problem described below:

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 maxn = 333, mod = 998244353;
bitset<maxn> e[maxn];
int power(int x){
	int v = 1;
	for(int i = 0; i < x; i++){
		v *= 2;
		if(v >= mod) v -= mod;
	}
	return v;
}
int main(){
	int n, m; cin >> n >> m; int rnk = 0;
	for(int i = 0; i < n; i++){
		bitset<maxn> now;
		for(int j = 0; j < m; j++){
			int v; cin >> v;
			now[j] = v == 1;
		}
		for(int j = 0; j < m; j++){
			if(e[j][j] && now[j]){
				now ^= e[j];
			} else if(now[j]){
				e[j] = now;
				rnk++;
				break;
			}
		}
	}
	int ans = power(n + m - 1);
	ans += mod - power(n + m - 1 - rnk);
	if(ans >= mod) ans -= mod;
	cout << ans << endl;
	return 0;
}


```