#include <bits/stdc++.h>
#define Pct puts("------ Yes ------");
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define per(a, b, c) for(int a = (b); a >= (c); -- a)
#define fir first
#define sec second
#define ls son[u][0]
#define rs son[u][1]
#define eps 1e-8
#define PI acos(-1.0)
using namespace std;
template <class T> void rd(T &x){ x = 0; int f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar(); x *= f; }

typedef long long ll;
const int maxn = 305;
const int mod = 998244353;
int n, m;
int mat[maxn][maxn];

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

int gauss(int n, int m){
	int r = 1, c = 1;
	for(; r <= n && c <= m; r ++, c ++){
		int p = -1;
		rep(i, r, n) if(mat[i][c]){ p = i; break; }
		if(p == -1){ r --; continue; }
		if(p != r) swap(mat[r], mat[p]);
		rep(i, r + 1, n) if(mat[i][c])
			per(j, m, c) mat[i][j] ^= mat[r][j];
	}
	return -- r;
}

int main(){
	rd(n), rd(m);
	rep(i, 1, n) rep(j, 1, m) rd(mat[i][j]);
	printf("%d\n", (qpow(2, n + m - 1) - qpow(2, n + m - 1 - gauss(n, m)) + mod) % mod);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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>
#define Pct puts("------ Yes ------");
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define per(a, b, c) for(int a = (b); a >= (c); -- a)
#define fir first
#define sec second
#define ls son[u][0]
#define rs son[u][1]
#define eps 1e-8
#define PI acos(-1.0)
using namespace std;
template <class T> void rd(T &x){ x = 0; int f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar(); x *= f; }

typedef long long ll;
const int maxn = 305;
const int mod = 998244353;
int n, m;
int mat[maxn][maxn];

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

int gauss(int n, int m){
	int r = 1, c = 1;
	for(; r <= n && c <= m; r ++, c ++){
		int p = -1;
		rep(i, r, n) if(mat[i][c]){ p = i; break; }
		if(p == -1){ r --; continue; }
		if(p != r) swap(mat[r], mat[p]);
		rep(i, r + 1, n) if(mat[i][c])
			per(j, m, c) mat[i][j] ^= mat[r][j];
	}
	return -- r;
}

int main(){
	rd(n), rd(m);
	rep(i, 1, n) rep(j, 1, m) rd(mat[i][j]);
	printf("%d\n", (qpow(2, n + m - 1) - qpow(2, n + m - 1 - gauss(n, m)) + mod) % mod);
	return 0;
}
```