#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <bitset>
#define MOD 998244353
using namespace std;
typedef long long ll;

template <typename Tp> inline void getint(Tp &num){
	register int ch, neg = 0;
	while(!isdigit(ch = getchar())) if(ch == '-') neg = 1;
	num = ch & 15;
	while(isdigit(ch = getchar())) num = num * 10 + (ch & 15);
	if(neg) num = -num;
}

inline int hiku(int a, const int &b) {return (a -= b) < 0 ? a + MOD : a;}

inline ll fastpow(ll bas, ll ex){
	register ll res = 1; bas %= MOD;
	for(; ex; ex >>= 1, bas = bas * bas % MOD) if(ex & 1) res = res * bas % MOD;
	return res;
}

int N, M;
bitset<300> a[300], b[300];

inline int insert(bitset<300> val){
	for(register int i = M - 1; ~i; i--) if(val[i]){
		if(!b[i][i]) return b[i] = val, 1;
		val ^= b[i];
	}
	return 0;
}

int main(){
	getint(N), getint(M); int r = 0;
	for(register int i = 0; i < N; i++)
		for(register int j = 0, v; j < M; j++) getint(v), a[i][j] = v;
	for(register int i = 0; i < N; i++) r += insert(a[i]);
	return printf("%d\n", hiku(fastpow(2, N + M - 1), fastpow(2, N + M - r - 1))), 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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <bitset>
#define MOD 998244353
using namespace std;
typedef long long ll;

template <typename Tp> inline void getint(Tp &num){
	register int ch, neg = 0;
	while(!isdigit(ch = getchar())) if(ch == '-') neg = 1;
	num = ch & 15;
	while(isdigit(ch = getchar())) num = num * 10 + (ch & 15);
	if(neg) num = -num;
}

inline int hiku(int a, const int &b) {return (a -= b) < 0 ? a + MOD : a;}

inline ll fastpow(ll bas, ll ex){
	register ll res = 1; bas %= MOD;
	for(; ex; ex >>= 1, bas = bas * bas % MOD) if(ex & 1) res = res * bas % MOD;
	return res;
}

int N, M;
bitset<300> a[300], b[300];

inline int insert(bitset<300> val){
	for(register int i = M - 1; ~i; i--) if(val[i]){
		if(!b[i][i]) return b[i] = val, 1;
		val ^= b[i];
	}
	return 0;
}

int main(){
	getint(N), getint(M); int r = 0;
	for(register int i = 0; i < N; i++)
		for(register int j = 0, v; j < M; j++) getint(v), a[i][j] = v;
	for(register int i = 0; i < N; i++) r += insert(a[i]);
	return printf("%d\n", hiku(fastpow(2, N + M - 1), fastpow(2, N + M - r - 1))), 0;
}
```