#include <cstdio>
#include <algorithm>
#include <bitset>

typedef unsigned int ui;
const int MN = 2000;
const int Mod = 998244353;

int N, M, Cnt;
int A[MN][MN];

int Sz; ui *B[MN], *tmp;
#define set(b, i, x) ((x) ? (b[(i) >> 5] |= 1 << ((i) & 31)) : (b[(i) >> 5] &= ~(1 << ((i) & 31))))
#define val(b, i) (b[(i) >> 5] >> ((i) & 31) & 1)

int main() {
	scanf("%d%d", &N, &M);
	if (N > M) {
		for (int i = 0; i < N; ++i)
			for (int j = 0; j < M; ++j)
				scanf("%d", &A[i][j]);
	}
	else {
		for (int i = 0; i < N; ++i)
			for (int j = 0; j < M; ++j)
				scanf("%d", &A[j][i]);
		std::swap(N, M);
	}
	Sz = (M + 31) >> 5;
	for (int j = 0; j < M; ++j) B[j] = new ui[Sz];
	tmp = new ui[Sz];
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < M; ++j) set(tmp, j, A[i][j]);
		for (int j = M - 1; ~j; --j) if (val(tmp, j)) {
			if (val(B[j], j)) for (int k = 0; k < Sz; ++k) tmp[k] ^= B[j][k];
			else {
				for (int k = 0; k < Sz; ++k) B[j][k] = tmp[k];
				++Cnt;
				break;
			}
		}
	}
	long long Ans = 0;
	for (int i = 0, s = 1; i < N + M; ++i, s = (s << 1) % Mod) {
		if (i == N + M - Cnt - 1) Ans -= s;
		if (i == N + M - 1) Ans += s;
	}
	printf("%lld\n", (Ans % Mod + Mod) % Mod);
	for (int j = 0; j < M; ++j) delete [] B[j];
	delete [] tmp;
	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 <cstdio>
#include <algorithm>
#include <bitset>

typedef unsigned int ui;
const int MN = 2000;
const int Mod = 998244353;

int N, M, Cnt;
int A[MN][MN];

int Sz; ui *B[MN], *tmp;
#define set(b, i, x) ((x) ? (b[(i) >> 5] |= 1 << ((i) & 31)) : (b[(i) >> 5] &= ~(1 << ((i) & 31))))
#define val(b, i) (b[(i) >> 5] >> ((i) & 31) & 1)

int main() {
	scanf("%d%d", &N, &M);
	if (N > M) {
		for (int i = 0; i < N; ++i)
			for (int j = 0; j < M; ++j)
				scanf("%d", &A[i][j]);
	}
	else {
		for (int i = 0; i < N; ++i)
			for (int j = 0; j < M; ++j)
				scanf("%d", &A[j][i]);
		std::swap(N, M);
	}
	Sz = (M + 31) >> 5;
	for (int j = 0; j < M; ++j) B[j] = new ui[Sz];
	tmp = new ui[Sz];
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < M; ++j) set(tmp, j, A[i][j]);
		for (int j = M - 1; ~j; --j) if (val(tmp, j)) {
			if (val(B[j], j)) for (int k = 0; k < Sz; ++k) tmp[k] ^= B[j][k];
			else {
				for (int k = 0; k < Sz; ++k) B[j][k] = tmp[k];
				++Cnt;
				break;
			}
		}
	}
	long long Ans = 0;
	for (int i = 0, s = 1; i < N + M; ++i, s = (s << 1) % Mod) {
		if (i == N + M - Cnt - 1) Ans -= s;
		if (i == N + M - 1) Ans += s;
	}
	printf("%lld\n", (Ans % Mod + Mod) % Mod);
	for (int j = 0; j < M; ++j) delete [] B[j];
	delete [] tmp;
	return 0;
}
```