#include "bits/stdc++.h"
using namespace std;
bitset <333> a[333];
int n, m;
const int mod = 998244353;
long long pw[333];

int main(int argc, char const *argv[])
{
	scanf("%d %d", &n, &m);
	pw[0] = 1;
	for(int i = 1; i <= max(n, m); i++) {
		pw[i] = pw[i - 1] * 2;
		pw[i] %= mod;
	} 
	for(int i = 0; i < n; i++) {
		for(int j = 0; j < m; j++) {
			int x;
			scanf("%d", &x);
			a[i][j] = x;
		}
	}
	int ptr = 0;
	for(int j = 0; j < m; j++) {
		for(int i = ptr; i < n; i++) {
			if(a[i][j] == 1) {
				swap(a[ptr], a[i]);
				for(int x = 0; x < n; x++) {
					if(x != ptr && a[x][j] == 1) {
						a[x] ^= a[ptr];
					}
				}
				++ptr;
				break;
			}
		}
	}
	long long res = pw[m-1] * (pw[n] - pw[n - ptr]);
	res %= mod;
	if(res < 0) res += mod;
	printf("%lld\n", res);
	return 0;
}
### Prompt

Generate 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;
bitset <333> a[333];
int n, m;
const int mod = 998244353;
long long pw[333];

int main(int argc, char const *argv[])
{
	scanf("%d %d", &n, &m);
	pw[0] = 1;
	for(int i = 1; i <= max(n, m); i++) {
		pw[i] = pw[i - 1] * 2;
		pw[i] %= mod;
	} 
	for(int i = 0; i < n; i++) {
		for(int j = 0; j < m; j++) {
			int x;
			scanf("%d", &x);
			a[i][j] = x;
		}
	}
	int ptr = 0;
	for(int j = 0; j < m; j++) {
		for(int i = ptr; i < n; i++) {
			if(a[i][j] == 1) {
				swap(a[ptr], a[i]);
				for(int x = 0; x < n; x++) {
					if(x != ptr && a[x][j] == 1) {
						a[x] ^= a[ptr];
					}
				}
				++ptr;
				break;
			}
		}
	}
	long long res = pw[m-1] * (pw[n] - pw[n - ptr]);
	res %= mod;
	if(res < 0) res += mod;
	printf("%lld\n", res);
	return 0;
}
```