#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define MOD 998244353
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

int H, W;
int A[302][302];

int main()
{
	scanf("%d%d", &H, &W);
	for (int i = 0; i < H; ++i) {
		for (int j = 0; j < W; ++j) scanf("%d", &(A[i][j]));
	}
	int piv = 0;
	for (int i = 0; i < W; ++i) {
		int r = -1;
		for (int j = piv; j < H; ++j) {
			if (A[j][i] == 1) {
				r = j;
				break;
			}
		}
		if (r == -1) continue;
		if (r != piv) for (int j = 0; j < W; ++j) swap(A[r][j], A[piv][j]);
		for (int j = 0; j < H; ++j) if (j != piv && A[j][i] == 1) {
			for (int k = i; k < W; ++k) A[j][k] ^= A[piv][k];
		}
		++piv;
	}
	i64 ret = 1;
	for (int i = 0; i < H; ++i) ret = ret * 2 % MOD;
	i64 sub = 1;
	for (int i = 0; i < (H - piv); ++i) sub = sub * 2 % MOD;
	ret += MOD - sub;
	ret %= MOD;
	for (int i = 0; i < W - 1; ++i) ret = ret * 2 % MOD;
	printf("%lld\n", ret);
	return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define MOD 998244353
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

int H, W;
int A[302][302];

int main()
{
	scanf("%d%d", &H, &W);
	for (int i = 0; i < H; ++i) {
		for (int j = 0; j < W; ++j) scanf("%d", &(A[i][j]));
	}
	int piv = 0;
	for (int i = 0; i < W; ++i) {
		int r = -1;
		for (int j = piv; j < H; ++j) {
			if (A[j][i] == 1) {
				r = j;
				break;
			}
		}
		if (r == -1) continue;
		if (r != piv) for (int j = 0; j < W; ++j) swap(A[r][j], A[piv][j]);
		for (int j = 0; j < H; ++j) if (j != piv && A[j][i] == 1) {
			for (int k = i; k < W; ++k) A[j][k] ^= A[piv][k];
		}
		++piv;
	}
	i64 ret = 1;
	for (int i = 0; i < H; ++i) ret = ret * 2 % MOD;
	i64 sub = 1;
	for (int i = 0; i < (H - piv); ++i) sub = sub * 2 % MOD;
	ret += MOD - sub;
	ret %= MOD;
	for (int i = 0; i < W - 1; ++i) ret = ret * 2 % MOD;
	printf("%lld\n", ret);
	return 0;
}

```