#include <iostream>
#include <vector>

using namespace std;

const int MOD = 998244353;

int main(){
    int N, M;
    while(cin >> N >> M){
        vector<vector<int>> A(N, vector<int>(M, 0));
        for(auto& v : A){
            for(auto& t : v) cin >> t;
        }
        int c = 0;
        int r = 0;
        for(int i=0;i<min(N, M);i++){
            int pivot = i;
            while(c < M){
                pivot = i;
                for(int j=i+1;j<N;j++){
                    if(A[pivot][c] < A[j][c]){
                        pivot = j;
                    }
                }
                if(A[pivot][c] == 0){ c++; continue; }
                break;
            }
            if(c == M) break;
            r++;
            if(pivot != i){
                for(int j=0;j<M;j++) swap(A[i][j], A[pivot][j]);
            }
            for(int j=i+1;j<N;j++){
                if(A[j][c] != A[i][c]) continue;
                for(int k=c;k<M;k++){
                    A[j][k] = (A[j][k] + A[i][k])%2;
                }
            }
        }
        long long powM = 1, powN = 1;
        for(int i=0;i<M-1;i++) powM = (2*powM) % MOD;
        for(int i=0;i<N;i++) powN = (2*powN) % MOD;
        powN = (powN + MOD - 1) % MOD;
        long long sub = 1;
        for(int i=N-1;i>=r;i--){
            powN = (powN + MOD - sub) % MOD;
            sub = (2*sub) % MOD;
        }
        cout << (powN*powM) % MOD << endl;
    }
}
### 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 <iostream>
#include <vector>

using namespace std;

const int MOD = 998244353;

int main(){
    int N, M;
    while(cin >> N >> M){
        vector<vector<int>> A(N, vector<int>(M, 0));
        for(auto& v : A){
            for(auto& t : v) cin >> t;
        }
        int c = 0;
        int r = 0;
        for(int i=0;i<min(N, M);i++){
            int pivot = i;
            while(c < M){
                pivot = i;
                for(int j=i+1;j<N;j++){
                    if(A[pivot][c] < A[j][c]){
                        pivot = j;
                    }
                }
                if(A[pivot][c] == 0){ c++; continue; }
                break;
            }
            if(c == M) break;
            r++;
            if(pivot != i){
                for(int j=0;j<M;j++) swap(A[i][j], A[pivot][j]);
            }
            for(int j=i+1;j<N;j++){
                if(A[j][c] != A[i][c]) continue;
                for(int k=c;k<M;k++){
                    A[j][k] = (A[j][k] + A[i][k])%2;
                }
            }
        }
        long long powM = 1, powN = 1;
        for(int i=0;i<M-1;i++) powM = (2*powM) % MOD;
        for(int i=0;i<N;i++) powN = (2*powN) % MOD;
        powN = (powN + MOD - 1) % MOD;
        long long sub = 1;
        for(int i=N-1;i>=r;i--){
            powN = (powN + MOD - sub) % MOD;
            sub = (2*sub) % MOD;
        }
        cout << (powN*powM) % MOD << endl;
    }
}
```