#include<bits/stdc++.h>
using namespace std;
const long long mod=998244353;
long long n,m,a[333][333],pw[666],cnt,pos,cur,upper;
/*long long Guass(long long Row,long long Column)
{
    long long row=1,col=1,max_r;
    for(row=1;row<=Row&&col<=Column;row++,col++)
    { 
        max_r=row;
        for(long long i=row+1;i<=Row;i++)
            if(abs(a[i][col])>abs(a[max_r][col]))
                max_r=i;
        if(a[max_r][col]==0)
        {
            row--;
            continue;
        }
        if(max_r!=row)
            for(long long i=col;i<=Column;i++)
                swap(a[row][i],a[max_r][i]);

        for(long long i=row+1;i<=Row;i++)
        {
            if(a[i][col]!=0)
            {

                for(long long j=col;j<=Column;j++)
                    a[i][j]^=a[row][j];
            }
        }
    }
	return Column-row+1;
}*/
void gauss()
{
	upper=1;
	for (long long i=1;i<=m;i++)
	{
		pos=-1;
		for (long long j=upper;j<=n;j++)
		{
			if (a[j][i])
			{
				pos=j;
				break;
			}
		}
		if (~pos)
		{
			for (long long j=1;j<=m;j++) swap(a[upper][j],a[pos][j]);
			for (long long j=upper+1;j<=n;j++)
			{
				if (a[j][i])
				{
					for (long long k=i;k<=m;k++)
					{
						a[j][k]^=a[upper][k];
					}
				}
			}
			upper++;
		}
	}
}
int main()
{
	scanf("%lld%lld",&n,&m);
	for (long long i=1;i<=n;i++)
	{
		for (long long j=1;j<=m;j++)
		{
			scanf("%lld",&a[i][j]);
		}
	}
	pw[0]=1;
	for (long long i=1;i<=600;i++) pw[i]=pw[i-1]*2%mod;
	gauss();
	cur=pw[m-upper+1];
	printf("%lld\n",((pw[n+m-1]-pw[n-1]*cur)%mod+mod)%mod);
	return 0;
}
### Prompt

Your challenge is to write 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;
const long long mod=998244353;
long long n,m,a[333][333],pw[666],cnt,pos,cur,upper;
/*long long Guass(long long Row,long long Column)
{
    long long row=1,col=1,max_r;
    for(row=1;row<=Row&&col<=Column;row++,col++)
    { 
        max_r=row;
        for(long long i=row+1;i<=Row;i++)
            if(abs(a[i][col])>abs(a[max_r][col]))
                max_r=i;
        if(a[max_r][col]==0)
        {
            row--;
            continue;
        }
        if(max_r!=row)
            for(long long i=col;i<=Column;i++)
                swap(a[row][i],a[max_r][i]);

        for(long long i=row+1;i<=Row;i++)
        {
            if(a[i][col]!=0)
            {

                for(long long j=col;j<=Column;j++)
                    a[i][j]^=a[row][j];
            }
        }
    }
	return Column-row+1;
}*/
void gauss()
{
	upper=1;
	for (long long i=1;i<=m;i++)
	{
		pos=-1;
		for (long long j=upper;j<=n;j++)
		{
			if (a[j][i])
			{
				pos=j;
				break;
			}
		}
		if (~pos)
		{
			for (long long j=1;j<=m;j++) swap(a[upper][j],a[pos][j]);
			for (long long j=upper+1;j<=n;j++)
			{
				if (a[j][i])
				{
					for (long long k=i;k<=m;k++)
					{
						a[j][k]^=a[upper][k];
					}
				}
			}
			upper++;
		}
	}
}
int main()
{
	scanf("%lld%lld",&n,&m);
	for (long long i=1;i<=n;i++)
	{
		for (long long j=1;j<=m;j++)
		{
			scanf("%lld",&a[i][j]);
		}
	}
	pw[0]=1;
	for (long long i=1;i<=600;i++) pw[i]=pw[i-1]*2%mod;
	gauss();
	cur=pw[m-upper+1];
	printf("%lld\n",((pw[n+m-1]-pw[n-1]*cur)%mod+mod)%mod);
	return 0;
}
```