#include <iostream>
#include <vector>

using namespace std;

int main()
{
    int N, M;
    cin>>N>>M;
    vector<vector<int>> a(N, vector<int>(M));
    for (auto &l: a)
        for (int &c: l)
            cin>>c;

    for (int i=0; i<N; i++)
    {
        int px = -1, py;
        for (int j=0; j<M && px==-1; j++)
            for (int k=i; k<N && px==-1; k++)
                if (a[k][j] != 0)
                    px = j,
                    py = k;
        if (px==-1)
            break;
        a[i].swap(a[py]);
        for (int j=i+1; j<N; j++)
            if (a[j][px] != 0)
                for (int k=0; k<M; k++)
                    a[j][k] ^= a[i][k];
    }
    int r = 0;
    for (int i=0; i<N; i++)
        for (int j=0; j<M; j++)
            if (a[i][j] != 0)
                r = i+1;

    long long MOD = 998244353;

    long long call = 1;
    for (int i=0; i<M; i++)
        call = call*2%MOD;
    long long czero = 1;
    for (int i=0; i<M-r; i++)
        czero = czero*2%MOD;
    long long rall = 1;
    for (int i=0; i<N-1; i++)
        rall = rall*2%MOD;
    cout<<(call-czero+MOD)%MOD*rall%MOD<<endl;
}

### Prompt

Please formulate 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 <iostream>
#include <vector>

using namespace std;

int main()
{
    int N, M;
    cin>>N>>M;
    vector<vector<int>> a(N, vector<int>(M));
    for (auto &l: a)
        for (int &c: l)
            cin>>c;

    for (int i=0; i<N; i++)
    {
        int px = -1, py;
        for (int j=0; j<M && px==-1; j++)
            for (int k=i; k<N && px==-1; k++)
                if (a[k][j] != 0)
                    px = j,
                    py = k;
        if (px==-1)
            break;
        a[i].swap(a[py]);
        for (int j=i+1; j<N; j++)
            if (a[j][px] != 0)
                for (int k=0; k<M; k++)
                    a[j][k] ^= a[i][k];
    }
    int r = 0;
    for (int i=0; i<N; i++)
        for (int j=0; j<M; j++)
            if (a[i][j] != 0)
                r = i+1;

    long long MOD = 998244353;

    long long call = 1;
    for (int i=0; i<M; i++)
        call = call*2%MOD;
    long long czero = 1;
    for (int i=0; i<M-r; i++)
        czero = czero*2%MOD;
    long long rall = 1;
    for (int i=0; i<N-1; i++)
        rall = rall*2%MOD;
    cout<<(call-czero+MOD)%MOD*rall%MOD<<endl;
}

```