// by tmt514
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#define SZ(x) ((int)(x).size())
#define FOR(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); ++it)
using namespace std;
typedef long long LL;

const LL MOD = 998244353;
LL bigmod(LL a, LL n) {
  if(n==0) return 1;
  if(n==1) return a%MOD;
  return bigmod(a*a%MOD, n/2) * (n%2? a:1)%MOD;
}
int a[1005][1005] = {};
int main(void) {
  int n, m;
  cin >> n >> m;
  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
      cin >> a[i][j];
  int rank=0, w=0;
  if(n>m) {
    for(int i=0;i<n;i++)
      for(int j=i+1;j<n;j++)
        swap(a[i][j], a[j][i]);
    swap(n, m);
  }
  for(int i=0;i<m;i++) {
    int g=-1;
    for(int j=w;j<n;j++)
      if(a[j][i]==1) {
        g=j;
        break;
      }
    if(g>=0) {
      for(int k=0;k<m;k++)
        swap(a[w][k], a[g][k]);
      for(int j=w+1;j<n;j++) {
          if(a[j][i]==1)
        for(int k=0;k<m;k++)
          a[j][k] ^= a[w][k];
      }
      ++rank;
      ++w;
    }
  }
  
  LL t = bigmod(2, m-1) * (bigmod(2, n) - bigmod(2, n-rank)) % MOD;
  t=(t%MOD+MOD)%MOD;
  cout << t << endl;
  
      
  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
// by tmt514
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#define SZ(x) ((int)(x).size())
#define FOR(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); ++it)
using namespace std;
typedef long long LL;

const LL MOD = 998244353;
LL bigmod(LL a, LL n) {
  if(n==0) return 1;
  if(n==1) return a%MOD;
  return bigmod(a*a%MOD, n/2) * (n%2? a:1)%MOD;
}
int a[1005][1005] = {};
int main(void) {
  int n, m;
  cin >> n >> m;
  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
      cin >> a[i][j];
  int rank=0, w=0;
  if(n>m) {
    for(int i=0;i<n;i++)
      for(int j=i+1;j<n;j++)
        swap(a[i][j], a[j][i]);
    swap(n, m);
  }
  for(int i=0;i<m;i++) {
    int g=-1;
    for(int j=w;j<n;j++)
      if(a[j][i]==1) {
        g=j;
        break;
      }
    if(g>=0) {
      for(int k=0;k<m;k++)
        swap(a[w][k], a[g][k]);
      for(int j=w+1;j<n;j++) {
          if(a[j][i]==1)
        for(int k=0;k<m;k++)
          a[j][k] ^= a[w][k];
      }
      ++rank;
      ++w;
    }
  }
  
  LL t = bigmod(2, m-1) * (bigmod(2, n) - bigmod(2, n-rank)) % MOD;
  t=(t%MOD+MOD)%MOD;
  cout << t << endl;
  
      
  return 0;
}

```