#include <cstdio>
#include <bitset>
#define mod 998244353

typedef std::bitset<301> BS;

BS d[301];
BS a[301];

int qpow(int x, int p)
{
	int ans = 1;
	while(p)
	{
		if(p & 1) ans = 1ll * ans * x % mod;
		x = 1ll * x * x % mod; p >>= 1; 
	}
	return ans;
}

int main()
{
	int n, m; scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
		{
			int x; scanf("%d", &x);
			a[j][i] = x;
		}
	int cnt = 0;
	for(int i = 1; i <= m; i++)
		for(int j = n; j >= 1; j--)
		{
			if(a[i][j] == 1)
			{
				if(!d[j][j]) d[j] = a[i], cnt++;
				a[i] ^= d[j];
			}
		}
	printf("%lld\n", 1ll * qpow(2, n - 1) * (qpow(2, m) - qpow(2, m - cnt) + mod) % mod);
}

//对于一维的情况，如果有1，那么答案是2^(len-1)，否则为0
//考虑选出某几列，如果把这几列中每一行的数字xor起来后有至少一行为1，那么就有2^(n-1)的贡献
//把每一列看成一个数，题目转化成 问有多少个列的子集异或和不为0
//考虑异或和为0的子集个数，构造线性基，那么对每个不在线性基中的数，都可以选或不选(加上几个在线性基中的数都能xor出0)
//所以为2^(m-cnt)，cnt为在线性基中的数的个数
//ans = 2 ^ (n-1) * (2 ^ m - 2 ^ (m-cnt) ) 
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <cstdio>
#include <bitset>
#define mod 998244353

typedef std::bitset<301> BS;

BS d[301];
BS a[301];

int qpow(int x, int p)
{
	int ans = 1;
	while(p)
	{
		if(p & 1) ans = 1ll * ans * x % mod;
		x = 1ll * x * x % mod; p >>= 1; 
	}
	return ans;
}

int main()
{
	int n, m; scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
		{
			int x; scanf("%d", &x);
			a[j][i] = x;
		}
	int cnt = 0;
	for(int i = 1; i <= m; i++)
		for(int j = n; j >= 1; j--)
		{
			if(a[i][j] == 1)
			{
				if(!d[j][j]) d[j] = a[i], cnt++;
				a[i] ^= d[j];
			}
		}
	printf("%lld\n", 1ll * qpow(2, n - 1) * (qpow(2, m) - qpow(2, m - cnt) + mod) % mod);
}

//对于一维的情况，如果有1，那么答案是2^(len-1)，否则为0
//考虑选出某几列，如果把这几列中每一行的数字xor起来后有至少一行为1，那么就有2^(n-1)的贡献
//把每一列看成一个数，题目转化成 问有多少个列的子集异或和不为0
//考虑异或和为0的子集个数，构造线性基，那么对每个不在线性基中的数，都可以选或不选(加上几个在线性基中的数都能xor出0)
//所以为2^(m-cnt)，cnt为在线性基中的数的个数
//ans = 2 ^ (n-1) * (2 ^ m - 2 ^ (m-cnt) ) 
```