#include <stdio.h>
#include <algorithm>
#include <bitset>

const int MOD = 998244353;

int main() {
  int N, M; scanf("%d%d", &N, &M);
  std::bitset<300> A[300];
  for(int i = 0; i < N; ++i) {
    for(int j = 0; j < M; ++j) {
      int a; scanf("%d", &a);
      A[i][j] = a;
    }
  }
  int r = 0;
  for(int j = 0; j < M; ++j) {
    int min_i = N;
    for(int i = r; i < N; ++i) {
      if(A[i][j]) {
        min_i = i;
        break;
      }
    }
    if(min_i == N) continue;
    if(min_i > r) {
      A[r] ^= A[min_i];
    }
    for(int i = r + 1; i < N; ++i) {
      if(A[i][j]) {
        A[i] ^= A[r];
      }
    }
    ++r;
  }
  // fprintf(stderr, "r = %d\n", r);
  int count = 1;
  for(int i = 0; i < r; ++i) count = 2 * count % MOD;
  --count; if(count < 0) count += MOD;
  for(int i = r; i < N + M - 1; ++i) count = 2 * count % MOD;
  printf("%d\n", count);
  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 <stdio.h>
#include <algorithm>
#include <bitset>

const int MOD = 998244353;

int main() {
  int N, M; scanf("%d%d", &N, &M);
  std::bitset<300> A[300];
  for(int i = 0; i < N; ++i) {
    for(int j = 0; j < M; ++j) {
      int a; scanf("%d", &a);
      A[i][j] = a;
    }
  }
  int r = 0;
  for(int j = 0; j < M; ++j) {
    int min_i = N;
    for(int i = r; i < N; ++i) {
      if(A[i][j]) {
        min_i = i;
        break;
      }
    }
    if(min_i == N) continue;
    if(min_i > r) {
      A[r] ^= A[min_i];
    }
    for(int i = r + 1; i < N; ++i) {
      if(A[i][j]) {
        A[i] ^= A[r];
      }
    }
    ++r;
  }
  // fprintf(stderr, "r = %d\n", r);
  int count = 1;
  for(int i = 0; i < r; ++i) count = 2 * count % MOD;
  --count; if(count < 0) count += MOD;
  for(int i = r; i < N + M - 1; ++i) count = 2 * count % MOD;
  printf("%d\n", count);
  return 0;
}

```