#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#define MAXN 305
#define P 998244353
#define INF 0x3f3f3f3f
#define rint register int
#define LL long long
#define LD long double
using namespace std;

int n, m, siz, a[MAXN][MAXN], bas[MAXN][MAXN], temp[MAXN], vis[MAXN];

int ksm(int x, int y)
{
    int ans=1;
    while(y)
    {
        if(y&1) ans=1LL*ans*x%P;
        x=1LL*x*x%P; y>>=1;
    }
    return ans;
}

void insert(int a[])
{
    memcpy(temp, a, sizeof(temp));
    for(rint i=1; i<=m; ++i)
        if(temp[i])
        {
            if(!vis[i])
            {
                for(rint j=1; j<=m; ++j) bas[i][j]=temp[j];
                vis[i]=1;
                siz++;
                break;
            }
            for(rint j=1; j<=m; ++j) temp[j]^=bas[i][j];
        }
}

int main()
{
    scanf("%d%d", &n, &m);
    for(rint i=1; i<=n; ++i)
        for(rint j=1; j<=m; ++j) scanf("%d", &a[i][j]);
    for(rint i=1; i<=n; ++i) insert(a[i]);
    printf("%d\n", 1LL*(ksm(2, n)-ksm(2, n-siz)+P)*ksm(2, m-1)%P);
}
### Prompt

Develop a solution in Cpp 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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#define MAXN 305
#define P 998244353
#define INF 0x3f3f3f3f
#define rint register int
#define LL long long
#define LD long double
using namespace std;

int n, m, siz, a[MAXN][MAXN], bas[MAXN][MAXN], temp[MAXN], vis[MAXN];

int ksm(int x, int y)
{
    int ans=1;
    while(y)
    {
        if(y&1) ans=1LL*ans*x%P;
        x=1LL*x*x%P; y>>=1;
    }
    return ans;
}

void insert(int a[])
{
    memcpy(temp, a, sizeof(temp));
    for(rint i=1; i<=m; ++i)
        if(temp[i])
        {
            if(!vis[i])
            {
                for(rint j=1; j<=m; ++j) bas[i][j]=temp[j];
                vis[i]=1;
                siz++;
                break;
            }
            for(rint j=1; j<=m; ++j) temp[j]^=bas[i][j];
        }
}

int main()
{
    scanf("%d%d", &n, &m);
    for(rint i=1; i<=n; ++i)
        for(rint j=1; j<=m; ++j) scanf("%d", &a[i][j]);
    for(rint i=1; i<=n; ++i) insert(a[i]);
    printf("%d\n", 1LL*(ksm(2, n)-ksm(2, n-siz)+P)*ksm(2, m-1)%P);
}
```