#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

const ll MOD = 998244353;


ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}

const int MAXN = 310;

bitset<MAXN> bb[MAXN];

int n, m;

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n >> m;
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j) {
			int x;
			cin >> x;
			bb[i][j] = x;
		}
	int now = 0;
	for (int i = 0; i < m; ++i) {
		int fl = 0;
		for (int j = now; j < n; ++j)
			if (bb[j][i]) {
				fl = 1;
				swap(bb[j], bb[now]);
				break;
			}
		if (!fl)
			continue;
		for (int j = now + 1; j < n; ++j) {
			if (bb[j][i])
				bb[j] ^= bb[now];
		}
		++now;
	}
	ll go = (pw(2, n) + MOD - pw(2, n - now)) % MOD;
	cout << (go * pw(2, m - 1)) % MOD << "\n";
	return 0;
}



### Prompt

Please create a solution in CPP 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
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

const ll MOD = 998244353;


ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}

const int MAXN = 310;

bitset<MAXN> bb[MAXN];

int n, m;

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n >> m;
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j) {
			int x;
			cin >> x;
			bb[i][j] = x;
		}
	int now = 0;
	for (int i = 0; i < m; ++i) {
		int fl = 0;
		for (int j = now; j < n; ++j)
			if (bb[j][i]) {
				fl = 1;
				swap(bb[j], bb[now]);
				break;
			}
		if (!fl)
			continue;
		for (int j = now + 1; j < n; ++j) {
			if (bb[j][i])
				bb[j] ^= bb[now];
		}
		++now;
	}
	ll go = (pw(2, n) + MOD - pw(2, n - now)) % MOD;
	cout << (go * pw(2, m - 1)) % MOD << "\n";
	return 0;
}



```