#include <bits/stdc++.h>
using namespace std;
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0); 
	long long pow[610] = {}; 
	long long mod = 998244353; 
	pow[0] = 1; 
	for(int i = 1; i < 605; i++) pow[i] = pow[i-1] * 2 % mod; 
	int N, M; cin >> N >> M; 
	bool A[310][310] = {}; 
	for(int i = 0; i < N; i++){
		for(int j = 0; j < M; j++){
			cin >> A[i][j]; 
		}
	}
	int p = -1; 
	for(int j = 0; j < M; j++){
		for(int i = p + 1; i < N; i++){
			if(A[i][j]){
				swap(A[i], A[p + 1]); 
				p++; 
				for(int k = 0; k < N; k++){
					if(k == p) continue; 
					if(A[k][j]){
						for(int l = 0; l < M; l++) A[k][l] ^= A[p][l]; 
					}
				}
			}
		}
	}
	p++; 
	cout << (pow[p] + mod - 1) * pow[N + M - p - 1] % mod << "\n"; 
}
### 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;
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0); 
	long long pow[610] = {}; 
	long long mod = 998244353; 
	pow[0] = 1; 
	for(int i = 1; i < 605; i++) pow[i] = pow[i-1] * 2 % mod; 
	int N, M; cin >> N >> M; 
	bool A[310][310] = {}; 
	for(int i = 0; i < N; i++){
		for(int j = 0; j < M; j++){
			cin >> A[i][j]; 
		}
	}
	int p = -1; 
	for(int j = 0; j < M; j++){
		for(int i = p + 1; i < N; i++){
			if(A[i][j]){
				swap(A[i], A[p + 1]); 
				p++; 
				for(int k = 0; k < N; k++){
					if(k == p) continue; 
					if(A[k][j]){
						for(int l = 0; l < M; l++) A[k][l] ^= A[p][l]; 
					}
				}
			}
		}
	}
	p++; 
	cout << (pow[p] + mod - 1) * pow[N + M - p - 1] % mod << "\n"; 
}
```