#include"bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
constexpr long long mod = 998244353;
typedef pair<int, int> P;
typedef vector<int> vec;
typedef vector<vec> mat;

int kj[100005], kji[100005];
int modpow(int a, int x) {
	int res = 1;
	while (x) {
		if (x & 1)res = res*a%mod;
		a = a*a%mod;
		x >>= 1;
	}
	return res;
}

void setkj(int n) {
	kj[0] = 1;
	rep(i, n)kj[i + 1] = kj[i] * (i + 1) % mod;
	rep(i, n + 1)kji[i] = modpow(kj[i], mod - 2);
}
int comb(int r, int c) {
	if (c<0 || r<c)return 0;
	return kj[r] * kji[c] % mod*kji[r - c] % mod;
}
int GJ(mat &A) {
	int N = A.size(), M = A[0].size();
	int rnk = 0;
	rep(i, M) {
		int pivot = -1;
		for (int j = rnk; j < N; j++) {
			if (A[j][i]) {
				pivot = j; break;
			}
		}
		if (pivot == -1)continue;
		swap(A[pivot], A[rnk]);

		rep(j, N) {
			if (j != rnk&&A[j][i]) {
				rep(k, M) {
					A[j][k] ^= A[rnk][k];
				}
			}
		}
		rnk++;
	}

	int res = 0;
	rep(i, N) {
		rep(j, M) {
			if (A[i][j]) {
				res++; break;
			}
		}
	}
	return res;
}

signed main() {
	setkj(305);
	int n, m; cin >> n >> m;
	mat A(n);
	rep(i, n) {
		A[i].resize(m);
		rep(j, m)cin >> A[i][j];
	}
	int r = GJ(A);

	/*
	rep(i, n) {
		rep(j, m)cout << A[i][j]; cout << endl;
	}
	cout << r << endl;
	*/

	int ans = modpow(2, n - r)*modpow(2, m - r) % mod;
	int d = 0;
	for (int i = 1; i <= r; i += 2) {
		d += modpow(3, r - i)*comb(r, i) % mod;
		d %= mod;
	}
	cout << d*ans%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"bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
constexpr long long mod = 998244353;
typedef pair<int, int> P;
typedef vector<int> vec;
typedef vector<vec> mat;

int kj[100005], kji[100005];
int modpow(int a, int x) {
	int res = 1;
	while (x) {
		if (x & 1)res = res*a%mod;
		a = a*a%mod;
		x >>= 1;
	}
	return res;
}

void setkj(int n) {
	kj[0] = 1;
	rep(i, n)kj[i + 1] = kj[i] * (i + 1) % mod;
	rep(i, n + 1)kji[i] = modpow(kj[i], mod - 2);
}
int comb(int r, int c) {
	if (c<0 || r<c)return 0;
	return kj[r] * kji[c] % mod*kji[r - c] % mod;
}
int GJ(mat &A) {
	int N = A.size(), M = A[0].size();
	int rnk = 0;
	rep(i, M) {
		int pivot = -1;
		for (int j = rnk; j < N; j++) {
			if (A[j][i]) {
				pivot = j; break;
			}
		}
		if (pivot == -1)continue;
		swap(A[pivot], A[rnk]);

		rep(j, N) {
			if (j != rnk&&A[j][i]) {
				rep(k, M) {
					A[j][k] ^= A[rnk][k];
				}
			}
		}
		rnk++;
	}

	int res = 0;
	rep(i, N) {
		rep(j, M) {
			if (A[i][j]) {
				res++; break;
			}
		}
	}
	return res;
}

signed main() {
	setkj(305);
	int n, m; cin >> n >> m;
	mat A(n);
	rep(i, n) {
		A[i].resize(m);
		rep(j, m)cin >> A[i][j];
	}
	int r = GJ(A);

	/*
	rep(i, n) {
		rep(j, m)cout << A[i][j]; cout << endl;
	}
	cout << r << endl;
	*/

	int ans = modpow(2, n - r)*modpow(2, m - r) % mod;
	int d = 0;
	for (int i = 1; i <= r; i += 2) {
		d += modpow(3, r - i)*comb(r, i) % mod;
		d %= mod;
	}
	cout << d*ans%mod << endl;
}
```