#include<bits/stdc++.h>
using namespace std;
int mod = 998244353;
string XOR(string a, string b){
    for(int i = 0; i < b.size(); i++)
        a[i] ^= b[i];
    return a;
}
int po(int x){
    int ans = 1;
    for(int i = 0; i < x; i++){
        ans = ans + ans;
        if(ans >= mod) ans -= mod;
    }
    return ans;
}
int main(){
    int n, m;
    cin >> n >> m;
    int ans = 0;
    vector<string> v;
    string nul = "";
    for(int i = 0; i < m; i++)
        nul += (char)0;
    for(int i = 1; i <= n; i++){
        string x = "";
        for(int i = 0; i < m; i++){
            int t;
            cin >> t;
            x += t;
        }
        for(int j = v.size() - 1; j >= 0; j --){
            x = min(x, XOR(x, v[j]));
        }
        v.push_back(x);
        if(x == nul) ans++;
        sort(v.begin(), v.end());
    }
    cout << (po(n+m-1) - po(ans + m - 1) + mod) % mod << endl;

}

### 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 mod = 998244353;
string XOR(string a, string b){
    for(int i = 0; i < b.size(); i++)
        a[i] ^= b[i];
    return a;
}
int po(int x){
    int ans = 1;
    for(int i = 0; i < x; i++){
        ans = ans + ans;
        if(ans >= mod) ans -= mod;
    }
    return ans;
}
int main(){
    int n, m;
    cin >> n >> m;
    int ans = 0;
    vector<string> v;
    string nul = "";
    for(int i = 0; i < m; i++)
        nul += (char)0;
    for(int i = 1; i <= n; i++){
        string x = "";
        for(int i = 0; i < m; i++){
            int t;
            cin >> t;
            x += t;
        }
        for(int j = v.size() - 1; j >= 0; j --){
            x = min(x, XOR(x, v[j]));
        }
        v.push_back(x);
        if(x == nul) ans++;
        sort(v.begin(), v.end());
    }
    cout << (po(n+m-1) - po(ans + m - 1) + mod) % mod << endl;

}

```