#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <iostream>
#include <stdio.h>
#include <vector>
#include <functional>
#include <unordered_set>
#include <algorithm>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <queue>
#include <stack>
#include <bitset>

using namespace std;

const int N = 2e5 + 10;
const long long INF = 1LL << 60;

void xorVector(vector<char>& a, const vector<char>& b) {
	for (int i = 0; i < a.size(); ++i) {
		a[i] ^= b[i];
	}
}

const int MOD = 998244353;

int add(int a, int b) {
	a += b;
	if (a >= MOD)
		return a - MOD;
	return a;
}

int main()
{
	int n, m;
	cin >> n >> m;
	vector <vector<char>> v(n);
	for (int i = 0; i < n; ++i) {
		v[i].resize(m);
		for (int j = 0; j < m; ++j) {
			cin >> v[i][j];
			v[i][j] = (v[i][j] == '1');
		}
	}
	int r = 0;
	for (int i = 0; i < m; ++i) {
		int id = -1;
		for (int j = r; j < n; ++j) {
			if (v[j][i] == 1) {
				id = j;
				break;
			}
		}
		if (id == -1) {
			continue;
		}
		v[r].swap(v[id]);
		for (int j = r + 1; j < n; ++j) {
			if (v[j][i]) {
				xorVector(v[j], v[r]);
			}
		}
		++r;
	}
	int pw2 = 1;
	vector <int> pow2 = { 1 };
	for (int i = 1; i <= n + m; ++i) {
		pw2 = add(pw2, pw2);
		pow2.push_back(pw2);
	}
	int ans = pow2[n + m - 1] - pow2[n + m - r - 1];
	if (ans < 0) {
		ans += MOD;
	}
	cout << ans << "\n";
	return 0;
}
### Prompt

Create a solution in cpp for 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
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <iostream>
#include <stdio.h>
#include <vector>
#include <functional>
#include <unordered_set>
#include <algorithm>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <queue>
#include <stack>
#include <bitset>

using namespace std;

const int N = 2e5 + 10;
const long long INF = 1LL << 60;

void xorVector(vector<char>& a, const vector<char>& b) {
	for (int i = 0; i < a.size(); ++i) {
		a[i] ^= b[i];
	}
}

const int MOD = 998244353;

int add(int a, int b) {
	a += b;
	if (a >= MOD)
		return a - MOD;
	return a;
}

int main()
{
	int n, m;
	cin >> n >> m;
	vector <vector<char>> v(n);
	for (int i = 0; i < n; ++i) {
		v[i].resize(m);
		for (int j = 0; j < m; ++j) {
			cin >> v[i][j];
			v[i][j] = (v[i][j] == '1');
		}
	}
	int r = 0;
	for (int i = 0; i < m; ++i) {
		int id = -1;
		for (int j = r; j < n; ++j) {
			if (v[j][i] == 1) {
				id = j;
				break;
			}
		}
		if (id == -1) {
			continue;
		}
		v[r].swap(v[id]);
		for (int j = r + 1; j < n; ++j) {
			if (v[j][i]) {
				xorVector(v[j], v[r]);
			}
		}
		++r;
	}
	int pw2 = 1;
	vector <int> pow2 = { 1 };
	for (int i = 1; i <= n + m; ++i) {
		pw2 = add(pw2, pw2);
		pow2.push_back(pw2);
	}
	int ans = pow2[n + m - 1] - pow2[n + m - r - 1];
	if (ans < 0) {
		ans += MOD;
	}
	cout << ans << "\n";
	return 0;
}
```