#include "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"
#include "functional"
#include "ctime"

using namespace std;

//constexpr long long int MOD = 1000000007;
//constexpr int MOD = 1000000007;
constexpr int MOD = 998244353;
//constexpr long long int MOD = 998244353;
constexpr double EPS = 1e-8;

//int N, M, K, H, W, L, R;
long long int N, M, K, H, W, L, R;

long long int power(long long int x, long long int n, long long int M) {
	long long int ret = 1;
	long long int by = x;
	while (n) {
		if (n & 1) {
			ret *= by;
			ret %= M;
		}
		by *= by;
		by %= M;
		n >>= 1;
	}
	return ret;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> H >> W;
	vector<bitset<300>>v(H);
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			cin >> K;
			if (K)v[i].flip(j);
		}
	}
	vector<int>used(H);
	for (int i = 0; i < W; i++) {
		for (int j = 0; j < H; j++) {
			if (used[j])continue;
			if (!v[j].test(i))continue;
			used[j] = 1;
			for (int k = 0; k < H; k++) {
				if (k == j)continue;
				if (v[k].test(i)) {
					v[k] ^= v[j];
				}
			}
		}
	}
	K = accumulate(used.begin(), used.end(), 0);
	cout << (power(2, K, MOD) - 1) * power(2, W - 1, MOD) % MOD*power(2, H - K, MOD) % MOD << endl;
}
### 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 "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"
#include "functional"
#include "ctime"

using namespace std;

//constexpr long long int MOD = 1000000007;
//constexpr int MOD = 1000000007;
constexpr int MOD = 998244353;
//constexpr long long int MOD = 998244353;
constexpr double EPS = 1e-8;

//int N, M, K, H, W, L, R;
long long int N, M, K, H, W, L, R;

long long int power(long long int x, long long int n, long long int M) {
	long long int ret = 1;
	long long int by = x;
	while (n) {
		if (n & 1) {
			ret *= by;
			ret %= M;
		}
		by *= by;
		by %= M;
		n >>= 1;
	}
	return ret;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> H >> W;
	vector<bitset<300>>v(H);
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			cin >> K;
			if (K)v[i].flip(j);
		}
	}
	vector<int>used(H);
	for (int i = 0; i < W; i++) {
		for (int j = 0; j < H; j++) {
			if (used[j])continue;
			if (!v[j].test(i))continue;
			used[j] = 1;
			for (int k = 0; k < H; k++) {
				if (k == j)continue;
				if (v[k].test(i)) {
					v[k] ^= v[j];
				}
			}
		}
	}
	K = accumulate(used.begin(), used.end(), 0);
	cout << (power(2, K, MOD) - 1) * power(2, W - 1, MOD) % MOD*power(2, H - K, MOD) % MOD << endl;
}
```