// ※※※ 解答不能 ※※※
// LHiC氏.
// https://atcoder.jp/contests/yahoo-procon2019-qual/submissions/4214380
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define rep(i, n) for(int i = 0; i < n; i++)
#define repx(i, a, b) for(int i = a; i < b; i++)
const LL MOD = 998244353;
const int MAXN = 310;
bitset<MAXN> bb[MAXN];

LL pw(LL a, LL b){
    LL ans = 1;
    while(b){
        while(!(b & 1)) b >>= 1, a = (a * a) % MOD;
        ans = (ans * a) % MOD, --b;
    }
    return ans;
}

int main(){
    int n, m, x;
    scanf("%d %d", &n, &m);
    rep(i, n){
        rep(j, m){
            scanf("%d", &x);
            bb[i][j] = x;
        }
    }
    int now = 0;
    rep(i, m){
        int fl = 0;
        repx(j, now, n){
            if(bb[j][i]){
                fl = 1;
                swap(bb[j], bb[now]);
                break;
            }
        }
        if(!fl) continue;
        repx(j, now + 1, n) if(bb[j][i]) bb[j] ^= bb[now];
        ++now;
    }
    LL go = (pw(2, n) + MOD - pw(2, n - now)) % MOD;
    printf("%lld\n", (go * pw(2, m - 1)) % MOD);
    return 0;
}
### 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
// ※※※ 解答不能 ※※※
// LHiC氏.
// https://atcoder.jp/contests/yahoo-procon2019-qual/submissions/4214380
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define rep(i, n) for(int i = 0; i < n; i++)
#define repx(i, a, b) for(int i = a; i < b; i++)
const LL MOD = 998244353;
const int MAXN = 310;
bitset<MAXN> bb[MAXN];

LL pw(LL a, LL b){
    LL ans = 1;
    while(b){
        while(!(b & 1)) b >>= 1, a = (a * a) % MOD;
        ans = (ans * a) % MOD, --b;
    }
    return ans;
}

int main(){
    int n, m, x;
    scanf("%d %d", &n, &m);
    rep(i, n){
        rep(j, m){
            scanf("%d", &x);
            bb[i][j] = x;
        }
    }
    int now = 0;
    rep(i, m){
        int fl = 0;
        repx(j, now, n){
            if(bb[j][i]){
                fl = 1;
                swap(bb[j], bb[now]);
                break;
            }
        }
        if(!fl) continue;
        repx(j, now + 1, n) if(bb[j][i]) bb[j] ^= bb[now];
        ++now;
    }
    LL go = (pw(2, n) + MOD - pw(2, n - now)) % MOD;
    printf("%lld\n", (go * pw(2, m - 1)) % MOD);
    return 0;
}
```