Problem: There was a girl who had a hobby of Othello. I enjoy Othello with my family when I have time.

One day, the girl tried to play Othello as usual. However, everyone was so busy that no one played. So I came up with a game that can be played by one person using Othello's board and spinning top.

problem

Othello pieces are arranged so that the top surface is black or white in all the squares of the vertical \\ (R \\) ​​square and the horizontal \\ (C \\) square.

Consider repeating the operation of "selecting a certain square and turning over all the pieces in the vertical, horizontal, and diagonal directions" as shown in the figure below. Find the number of operations that make the tops of all the pieces white. However, the following conditions must be met.

* You cannot operate the same square twice
* Do not distinguish the order of operations
* It is possible to continue the operation after the top surface of all the pieces has turned white.



<image>

input

The integer \\ (R \\) ​​and the integer \\ (C \\) are given on the first line, separated by blanks.

Subsequently, \\ (R \\) ​​lines are given \\ (C \\) integer values ​​(0 or 1), respectively. The \\ (c \\) th integer on the \\ (r \\) line represents the state of the square (r, c).

1 indicates that the upper surface is black, and 0 indicates that the upper surface is white.

output

If you can make the top surface of all the pieces on the board white, output the remainder of dividing the number of such operations by \\ (1000000009 (= 10 ^ 9 + 9) \\) on one line.

If you can't make the tops of all the pieces on the board white, print \\ (0 \\) on one line.

Constraint

* \\ (1 \ leq R \ leq 50, 1 \ leq C \ leq 50 \\)



Sample input / output

Input 1


13
1 1 1


Output 1


Four


\\ (\\ {(1,1) \\}, \\ {(1,2) \\}, \\ {(1,3) \\}, \\ {(1,1), (1) , 2), (1,3) \\} \\).

Input 2


13
1 0 1


Output 2


0


Input 3


twenty four
0 1 0 1
1 0 1 0


Output 3


1






Example

Input

1 3
1 1 1


Output

4

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <bitset>
using namespace std;
typedef long long ll;

// binary行列の掃き出し法と行列累乗
// bitset高速化により、定数倍速い

const int MAX_ROW = 2500;
const int MAX_COL = 2500;
struct BitMatrix {
    int H, W;
    bitset<MAX_COL> val[MAX_ROW];
    BitMatrix(int m, int n) : H(m), W(n) {}
    inline bitset<MAX_COL> & operator [] (int i) { return val[i]; }
};
inline BitMatrix operator * (BitMatrix A, BitMatrix B) {
    BitMatrix R(A.H, B.W);
    BitMatrix tB(B.W, B.H);
    for (int i = 0; i < tB.H; i++) for (int j = 0; j < tB.W; j++) tB[i][j] = B[j][i];
    for (int i = 0; i < R.H; i++) for (int j = 0; j < R.W; j++) R[i][j] = ((A[i] & tB[j]).count() & 1);
    return R;
}
inline BitMatrix pow(BitMatrix A, ll  n) {
    BitMatrix R(A.H, A.H);
    for (int i = 0; i < A.H; ++i) R[i][i] = 1;
    while (n > 0) { if (n & 1) R = R * A; A = A * A; n >>= 1; }
    return R;
}
int GaussJordan(BitMatrix &A, bool is_extended = false) {
    int rank = 0;
    for (int col = 0; col < A.W; col++) {
        if (is_extended && col == A.W - 1) break; // 拡大係数行列。この場合、最後の列は掃き出さない
        int pivot = -1;
        for (int row = rank; row < A.H; row++) if (A[row][col]) { pivot = row; break; }
        if (pivot == -1) continue;
        swap(A[pivot], A[rank]);
        for (int row = 0; row < A.H; row++) if (row != rank && A[row][col]) A[row] ^= A[rank];
        rank++;
    }
    return rank;
}
int linear_equation(BitMatrix A, vector<int> b, vector<int> &res) {
    int m = A.H, n = A.W;
    BitMatrix M(m, n + 1);
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) M[i][j] = A[i][j];
        M[i][n] = b[i];
    }
    int rank = GaussJordan(M, true);
    for (int row = rank; row < m; row++) if (M[row][n]) return -1; // check if it has no solution
    res.assign(n, 0);
    for (int i = 0; i < rank; i++) res[i] = M[i][n];
    return rank;
}

ll pow_mod(ll num, ll pow, ll mod) {
    ll prod = 1;
    num %= mod;
    while (pow > 0) {
        if (pow & 1) prod = prod * num % mod;
        num = num * num % mod;
        pow >>= 1;
    }
    return prod;
}


// verified
//  https://atcoder.jp/contests/code-thanks-festival-2017/tasks/code_thanks_festival_2017_f

// verified
//  https://atcoder.jp/contests/abc141/tasks/abc141_f

// verified
//  https://yukicoder.me/problems/no/184

// verified
//  https://yukicoder.me/problems/no/803

// verified
//  http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1308&lang=en
void AOJ_Awkward_Lights() {
    int m, n, d;
    while (true) {
        cin >> m >> n >> d;
        if (m == 0) break;
        vector <vector<int>> s(n, vector<int>(m));
        for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) cin >> s[i][j];
        BitMatrix A(n * m, n * m);
        vector<int> b(n * m), res;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                A[i * m + j][i * m + j] = 1;
                b[i * m + j] = 1 - s[i][j];
                for (int k = 0; k < n; k++)
                    for (int l = 0; l < m; l++)
                        if (abs(i - k) + abs(j - l) == d) A[i * m + j][k * m + l] = 1;
            }
        }
        cout << (linear_equation(A, b, res) == -1 ? 0 : 1) << '\n';
    }
}


// verified
//  http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2530
void AOJ_Reverse_Game() {
    int r, c;
    cin >> r >> c;
    vector<int> b(r * c), res;
    for (int i = 0; i < r; i++) for (int j = 0; j < c; j++) cin >> b[i * c + j];
    BitMatrix A(r * c, r * c);
    for (int i = 0; i < r; i++)
        for (int j = 0; j < c; j++)
            for (int k = 0; k < r; k++)
                for (int l = 0; l < c; l++)
                    if (i == k || j == l || abs(i - k) == abs(j - l))
                        A[i * c + j][k * c + l] = 1;
    int rank = linear_equation(A, b, res);
    cout << (rank == -1 ? 0 : pow_mod(2, r * c - rank, 1000000009)) << '\n';
}

int main() {
    // AOJ_Awkward_Lights();
    AOJ_Reverse_Game();
    return 0;
}