#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

typedef int _loop_int;
#define REP(i,n) for(_loop_int i=0;i<(_loop_int)(n);++i)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1;i>=(_loop_int)(a);--i)

#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define DEBUG_VEC(v) cout<<#v<<":";REP(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

#define CHMIN(a,b) a=min((a),(b))
#define CHMAX(a,b) a=max((a),(b))

// mod
const ll MOD = 998244353ll;

typedef bitset<300> bit;

int n,m;
bit a[325];

ll modpow(ll a, ll b){
  ll r = 1;
  while(b){
    if(b&1)r=r*a%MOD;
    a=a*a%MOD;
    b>>=1;
  }
  return r;
}

int main(){
  scanf("%d%d",&n,&m);
  REP(i,n){
    int x;
    REP(j,m){
      scanf("%d",&x);
      if(x==1)a[i].set(j);
    }
  }
  int cnt = 0;
  REP(pos,m){
    int tgt = -1;
    FOR(i,cnt,n)if(a[i][pos]){
      tgt = i; break;
    }
    if(tgt==-1)continue;
    swap(a[cnt], a[tgt]);
    REP(i,n)if(i != cnt && a[i][pos]){
      a[i] ^= a[cnt];
    }
    cnt++;
  }
  ll ans = (modpow(2ll, cnt) - 1 + MOD) * modpow(2ll, n-cnt) % MOD * modpow(2ll, m-1) % MOD;
  printf("%lld\n",ans);
  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;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

typedef int _loop_int;
#define REP(i,n) for(_loop_int i=0;i<(_loop_int)(n);++i)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1;i>=(_loop_int)(a);--i)

#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define DEBUG_VEC(v) cout<<#v<<":";REP(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

#define CHMIN(a,b) a=min((a),(b))
#define CHMAX(a,b) a=max((a),(b))

// mod
const ll MOD = 998244353ll;

typedef bitset<300> bit;

int n,m;
bit a[325];

ll modpow(ll a, ll b){
  ll r = 1;
  while(b){
    if(b&1)r=r*a%MOD;
    a=a*a%MOD;
    b>>=1;
  }
  return r;
}

int main(){
  scanf("%d%d",&n,&m);
  REP(i,n){
    int x;
    REP(j,m){
      scanf("%d",&x);
      if(x==1)a[i].set(j);
    }
  }
  int cnt = 0;
  REP(pos,m){
    int tgt = -1;
    FOR(i,cnt,n)if(a[i][pos]){
      tgt = i; break;
    }
    if(tgt==-1)continue;
    swap(a[cnt], a[tgt]);
    REP(i,n)if(i != cnt && a[i][pos]){
      a[i] ^= a[cnt];
    }
    cnt++;
  }
  ll ans = (modpow(2ll, cnt) - 1 + MOD) * modpow(2ll, n-cnt) % MOD * modpow(2ll, m-1) % MOD;
  printf("%lld\n",ans);
  return 0;
}

```