#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 300, md = 998244353;
int n, m, p2[N + 1];
vector<bitset<N> > g;

void ad(bitset<N> v){
	int j = 0;
	f(i, 0, g.size()){
		if (v.none())return;
		while (!v[j] && !g[i][j])++j;
		if (g[i][j] && v[j])v ^= g[i];
		else if (v[j]){
			g.insert(g.begin() + i, v);
			return;
		}
	}
	if (v.any())g.push_back(v);
}

int main(){
	p2[0] = 1;
	f(i, 1, N + 1)p2[i] = (ll)p2[i - 1] * 2 % md;
	scanf("%d%d", &n, &m);
	f(i, 0, n){
		bitset<N> v;
		f(j, 0, m){
			int t;
			scanf("%d", &t);
			if (t)v[j] = true;
		}
		ad(v);
	}
	int an = p2[n] - p2[n - g.size()] + md;
	an = (ll)an * p2[m - 1] % md;
	printf("%d\n", an);
}
### 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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 300, md = 998244353;
int n, m, p2[N + 1];
vector<bitset<N> > g;

void ad(bitset<N> v){
	int j = 0;
	f(i, 0, g.size()){
		if (v.none())return;
		while (!v[j] && !g[i][j])++j;
		if (g[i][j] && v[j])v ^= g[i];
		else if (v[j]){
			g.insert(g.begin() + i, v);
			return;
		}
	}
	if (v.any())g.push_back(v);
}

int main(){
	p2[0] = 1;
	f(i, 1, N + 1)p2[i] = (ll)p2[i - 1] * 2 % md;
	scanf("%d%d", &n, &m);
	f(i, 0, n){
		bitset<N> v;
		f(j, 0, m){
			int t;
			scanf("%d", &t);
			if (t)v[j] = true;
		}
		ad(v);
	}
	int an = p2[n] - p2[n - g.size()] + md;
	an = (ll)an * p2[m - 1] % md;
	printf("%d\n", an);
}
```