#include <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const ll mod=998244353;
const int maxn=310;
int n,m,cnt;
ll mi[maxn],ans;
bitset<maxn> a,p[maxn];
int main() {
	//freopen("1.txt","r",stdin);
	read(n); read(m); mi[0]=1; int x;
	for (int i=1;i<=max(n,m);i++) mi[i]=mi[i-1]*2%mod;
	for (int i=1;i<=n;i++) {
		for (int j=0;j<m;j++) read(x),a.set(j,x);
		for (int j=m-1;j>=0;j--) {
			if (a[j]==0) continue;
			if (p[j].none()) { p[j]=a; cnt++; break; }
			a^=p[j];
		}
	}
	ans=(mi[n]-mi[n-cnt]+mod)*mi[m-1]%mod;
	printf("%lld\n",ans);
	return 0;
}
### 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 <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const ll mod=998244353;
const int maxn=310;
int n,m,cnt;
ll mi[maxn],ans;
bitset<maxn> a,p[maxn];
int main() {
	//freopen("1.txt","r",stdin);
	read(n); read(m); mi[0]=1; int x;
	for (int i=1;i<=max(n,m);i++) mi[i]=mi[i-1]*2%mod;
	for (int i=1;i<=n;i++) {
		for (int j=0;j<m;j++) read(x),a.set(j,x);
		for (int j=m-1;j>=0;j--) {
			if (a[j]==0) continue;
			if (p[j].none()) { p[j]=a; cnt++; break; }
			a^=p[j];
		}
	}
	ans=(mi[n]-mi[n-cnt]+mod)*mi[m-1]%mod;
	printf("%lld\n",ans);
	return 0;
}
```