#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_set>
#include <bitset>
#include <time.h>
#include <limits.h>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define szz(x) (int)x.size()
#define sz(x) (int)x.size()
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) x.begin(),x.end()
typedef tuple<int, int, int> t3;

int n, m;
bitset <310> B[310];

ll pw[1010];
const ll mod =  998244353;

int main() {
	pw[0] = 1;
	for(int i=1;i<1010;i++) pw[i] = pw[i-1] * 2 % mod;
	scanf("%d%d", &n, &m);
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=m;j++) {
			int x; scanf("%d", &x);
			if(x & 1) B[i][j] = 1;
		}
	}
	int t = 1;
	for(int i=1;i<=m;i++) {
		int f = -1;
		for(int j=t;j<=n;j++) if(B[j][i]) {
			f = j; break;
		}
		if(f == -1) continue;
		swap(B[t], B[f]);
		for(int j=t+1;j<=n;j++) {
			if(B[j][i]) B[j] ^= B[t];
		}
		++t;
	}
	int r = t - 1;
	ll ans = (pw[n+m-1] - pw[n+m-1-r] + mod) % mod;
	printf("%lld\n", ans);
	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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_set>
#include <bitset>
#include <time.h>
#include <limits.h>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define szz(x) (int)x.size()
#define sz(x) (int)x.size()
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) x.begin(),x.end()
typedef tuple<int, int, int> t3;

int n, m;
bitset <310> B[310];

ll pw[1010];
const ll mod =  998244353;

int main() {
	pw[0] = 1;
	for(int i=1;i<1010;i++) pw[i] = pw[i-1] * 2 % mod;
	scanf("%d%d", &n, &m);
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=m;j++) {
			int x; scanf("%d", &x);
			if(x & 1) B[i][j] = 1;
		}
	}
	int t = 1;
	for(int i=1;i<=m;i++) {
		int f = -1;
		for(int j=t;j<=n;j++) if(B[j][i]) {
			f = j; break;
		}
		if(f == -1) continue;
		swap(B[t], B[f]);
		for(int j=t+1;j<=n;j++) {
			if(B[j][i]) B[j] ^= B[t];
		}
		++t;
	}
	int r = t - 1;
	ll ans = (pw[n+m-1] - pw[n+m-1-r] + mod) % mod;
	printf("%lld\n", ans);
	return 0;
}

```