#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int H,W;
ll mo=998244353;

const int MAT=402;
int A[402][402];
ll dp[603][603];
ll p2[3030];

int gf2_rank(int A[MAT][MAT]) { /* input */
	int i,j,k;
	FOR(i,H) {
		int be=i,mi=W+1;
		for(j=i;j<H;j++) {
			FOR(k,W) if(A[j][k]) break;
			if(k<mi) be=j,mi=k;
		}
		if(mi>=W) break;
		FOR(j,W) swap(A[i][j],A[be][j]);
		
		FOR(j,H) if(i!=j&&A[j][mi]) {
			FOR(k,W) A[j][k] ^= A[i][k];
		}
	}
	return i;
}


void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>H>>W;
	FOR(y,H) FOR(x,W) cin>>A[y][x];
	
	p2[0]=1;
	FOR(i,305) p2[i+1]=p2[i]*2%mo;
	
	x=gf2_rank(A);
	ll ret=(p2[x]+mo-1)*p2[H-x]%mo*p2[W-1]%mo;
	cout<<ret<<endl;
	
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}

### Prompt

Your task is to create 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;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int H,W;
ll mo=998244353;

const int MAT=402;
int A[402][402];
ll dp[603][603];
ll p2[3030];

int gf2_rank(int A[MAT][MAT]) { /* input */
	int i,j,k;
	FOR(i,H) {
		int be=i,mi=W+1;
		for(j=i;j<H;j++) {
			FOR(k,W) if(A[j][k]) break;
			if(k<mi) be=j,mi=k;
		}
		if(mi>=W) break;
		FOR(j,W) swap(A[i][j],A[be][j]);
		
		FOR(j,H) if(i!=j&&A[j][mi]) {
			FOR(k,W) A[j][k] ^= A[i][k];
		}
	}
	return i;
}


void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>H>>W;
	FOR(y,H) FOR(x,W) cin>>A[y][x];
	
	p2[0]=1;
	FOR(i,305) p2[i+1]=p2[i]*2%mo;
	
	x=gf2_rank(A);
	ll ret=(p2[x]+mo-1)*p2[H-x]%mo*p2[W-1]%mo;
	cout<<ret<<endl;
	
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}

```