#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define sz(x) (int)x.size()
#define mem(a) memset(a, 0, sizeof(a))
#define sqr(x) ((x) * (x))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int mn = 305;
const int mod = 998244353;
#define P mod

int n,m;

int b[mn][mn];

ll powmod(ll a, ll b) {
    ll t = 1;
    for (; b; b >>= 1, a = a*a%P)
        if (b & 1) t = t*a%P;
    return t;
}

bitset<300> p[mn],a[mn];

int r=0;

void init() {
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < m; j++) {
            if (!a[i][j]) continue;//对线性基的这一位没有贡献
            if (!p[j].any()) {
                p[j] = a[i];
                r++;
                break;
            }//选入线性基中
            a[i] ^= p[j];
        }
    }
}

int main() {
#ifdef LOCAL
    freopen("../1.txt", "r", stdin);
#endif
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        for(int j=0;j<m;j++){
              scanf("%d",&b[i][j]);
              a[i][j]=b[i][j];
        }
    }
    init();
    printf("%lld\n",(powmod(2,n+m-1)-powmod(2,n+m-r-1)+mod)%mod);
    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 pb push_back
#define mp make_pair
#define sz(x) (int)x.size()
#define mem(a) memset(a, 0, sizeof(a))
#define sqr(x) ((x) * (x))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int mn = 305;
const int mod = 998244353;
#define P mod

int n,m;

int b[mn][mn];

ll powmod(ll a, ll b) {
    ll t = 1;
    for (; b; b >>= 1, a = a*a%P)
        if (b & 1) t = t*a%P;
    return t;
}

bitset<300> p[mn],a[mn];

int r=0;

void init() {
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < m; j++) {
            if (!a[i][j]) continue;//对线性基的这一位没有贡献
            if (!p[j].any()) {
                p[j] = a[i];
                r++;
                break;
            }//选入线性基中
            a[i] ^= p[j];
        }
    }
}

int main() {
#ifdef LOCAL
    freopen("../1.txt", "r", stdin);
#endif
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        for(int j=0;j<m;j++){
              scanf("%d",&b[i][j]);
              a[i][j]=b[i][j];
        }
    }
    init();
    printf("%lld\n",(powmod(2,n+m-1)-powmod(2,n+m-r-1)+mod)%mod);
    return 0;
}

```