#include <bits/stdc++.h>

using namespace std;

#define MOD 998244353

int n, m, r, ans;
bitset<300> bm[300];
bool u[300];
int ta;

int po(int p) {
	long long re = 1, b = 2;
	while (p) {
		if (p & 1)
			re = re * b % MOD;
		b = b * b % MOD;
		p >>= 1;
	}
	return re;
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++) {
			scanf("%d", &ta);
			if (ta)
				bm[i].set(j);
		}
	for (int i = 0; i < m; i++) {
		ta = -1;
		for (int j = 0; j < n; j++)
			if (!u[j] && bm[j].test(i)) {
				u[j] = 1;
				ta = j;
				break;
			}
		if (ta == -1)
			continue;
		r++;
		for (int j = 0; j < n; j++)
			if (!u[j] && bm[j].test(i))
				bm[j] ^= bm[ta];
	}
	ans = po(n + m - 1) + MOD - po(n + m - r - 1);
	if (ans >= MOD)
		ans -= MOD;
	printf("%d\n", ans);
	return 0;
}
### Prompt

In cpp, your task is to solve 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
#include <bits/stdc++.h>

using namespace std;

#define MOD 998244353

int n, m, r, ans;
bitset<300> bm[300];
bool u[300];
int ta;

int po(int p) {
	long long re = 1, b = 2;
	while (p) {
		if (p & 1)
			re = re * b % MOD;
		b = b * b % MOD;
		p >>= 1;
	}
	return re;
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++) {
			scanf("%d", &ta);
			if (ta)
				bm[i].set(j);
		}
	for (int i = 0; i < m; i++) {
		ta = -1;
		for (int j = 0; j < n; j++)
			if (!u[j] && bm[j].test(i)) {
				u[j] = 1;
				ta = j;
				break;
			}
		if (ta == -1)
			continue;
		r++;
		for (int j = 0; j < n; j++)
			if (!u[j] && bm[j].test(i))
				bm[j] ^= bm[ta];
	}
	ans = po(n + m - 1) + MOD - po(n + m - r - 1);
	if (ans >= MOD)
		ans -= MOD;
	printf("%d\n", ans);
	return 0;
}
```