1. Abridged Problem Statement

There are N cells (1 ≤ N ≤ 200), each colored 0/1 (white/black).
For each cell i, you are given a set S_i of cells whose colors are toggled (XOR with 1) when you click i. You may click any cell any number of times (clicking twice cancels out).

Find any sequence of clicks that makes all cells the same color (all 0 or all 1).
If impossible, print -1. Otherwise print the number of clicks L and the list of clicked cell indices.

2. Detailed Editorial

Key observation: clicks are XOR / linear over GF(2)

Each cell’s color is a bit (0/1). Clicking a cell toggles some set of cells, i.e. adds 1 mod 2 to those positions. Therefore the entire process is linear over the field GF(2).

Because toggling twice cancels, for each cell i we only care whether we click it 0 or 1 times.
Let:
- x_i ∈ {0,1} = whether we click cell i.
- a_j ∈ {0,1} = initial color of cell j.

Let S_i be the set of cells toggled by clicking i. Then the final color of cell j is:
  a_j ⊕ (XOR over i with j ∈ S_i of x_i)

We want all final colors equal to some target t ∈ {0,1}:
  a_j ⊕ (XOR over i with j ∈ S_i of x_i) = t   for all j
Rearrange:
  XOR over i with j ∈ S_i of x_i = a_j ⊕ t

This is a system of N linear equations in N unknowns over GF(2).

Solve twice: target all-0 or all-1

We don’t know t. Try both:
- t=1 (make everything black)
- t=0 (make everything white)

For each target, build and solve the linear system; if solvable, output any solution.

Building the matrix

Define an N×N matrix M over GF(2) such that:
- M_{j,i} = 1 iff j ∈ S_i (click i affects cell j).
Then equations are M x = b where b_j = a_j ⊕ t.

Gaussian elimination over GF(2) with bitsets

N ≤ 200. Gaussian elimination is O(N^3), which is fine, but the time limit is tight (0.25s). Using bitset makes row XOR very fast (word-parallel), giving about O(N^3 / w).

Algorithm:
1. Each equation row is a bitset of length N+1: N coefficients plus 1 RHS bit.
2. For each column, find a pivot row with a 1, swap it into position, then XOR it into all other rows that have a 1 in that column (to zero them).
3. After elimination, if a row has all-zero coefficients but RHS = 1, the system is inconsistent.

Extracting a solution

This implementation produces a reduced form where each pivot column has its pivot row and other rows cleared in that column. A valid assignment is:
- Set all free variables to 0.
- For each pivot variable, its value is simply the RHS of its pivot row (since free vars are 0).

This gives “any solution”, which is acceptable.

Output

Collect all indices i where x_i = 1. Output count and list. If neither target works, output -1.

(Aside: if the cells formed an R×C grid, one only needs min(R, C) variables, since the first row/column determines all others. That optimization is not needed here.)

3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

template<size_t N>
class GaussBitset {
  public:
    int n_var;
    vector<bitset<N + 1>> rows;
    vector<int> where;
    bool solvable = false;

    GaussBitset(int n_var_) : n_var(n_var_) {}

    void add_equation(const vector<int>& coeffs, int rhs) {
        bitset<N + 1> row;
        for(int i = 0; i < n_var; i++) {
            if(coeffs[i] % 2) {
                row[i] = 1;
            }
        }
        row[n_var] = rhs % 2;
        rows.push_back(row);
    }

    void eliminate() {
        int n_eq = (int)rows.size();
        where.assign(n_var, -1);
        int row = 0;
        for(int col = 0; col < n_var && row < n_eq; col++) {
            int sel = -1;
            for(int i = row; i < n_eq; i++) {
                if(rows[i][col]) {
                    sel = i;
                    break;
                }
            }
            if(sel == -1) {
                continue;
            }
            swap(rows[sel], rows[row]);
            where[col] = row;
            for(int i = 0; i < n_eq; i++) {
                if(i != row && rows[i][col]) {
                    rows[i] ^= rows[row];
                }
            }
            row++;
        }
        solvable = true;
        for(int i = row; i < n_eq; i++) {
            if(rows[i][n_var]) {
                solvable = false;
                break;
            }
        }
    }

    bool has_solution() const { return solvable; }

    vector<int> free_variables() const {
        vector<int> fv;
        for(int v = 0; v < n_var; v++) {
            if(where[v] == -1) {
                fv.push_back(v);
            }
        }
        return fv;
    }

    vector<int> any_solution() const {
        vector<int> x(n_var, 0);
        for(int v = 0; v < n_var; v++) {
            if(where[v] != -1) {
                x[v] = rows[where[v]][n_var];
            }
        }
        return x;
    }
};

int n;
vector<vector<int>> groups;
vector<int> initial;

void read() {
    cin >> n;
    groups.assign(n, {});
    for(int i = 0; i < n; i++) {
        int k;
        cin >> k;
        groups[i].resize(k);
        for(auto& x: groups[i]) {
            cin >> x;
            x--;
        }
    }
    initial.assign(n, 0);
    for(auto& x: initial) {
        cin >> x;
    }
}

void solve() {
    // To solve this, we can solve the problem twice - once to try and activate
    // everything to be white, and once to try to have everything black. WLOG,
    // let's try to make everything black, and have black = 1, and white = 0.
    // Let's also have n variables x1, ..., xn, where xi = 1 if we activated a
    // cell. This means that we have N linear equations defined as x[i] +
    // sum(x[neighbours(i)]) = 1 mod 2. This can be solved with Gauss in O(N^3),
    // and optimized with bitsets further to O(N^3/w). We implement the
    // optimized solution here.
    //
    // Something interesting, but less relevant for the problem: if the problem
    // was over a R x C grid, we can actually only create min(R, C) variables
    // rather than the full R*C, because the first row/column in a grid implies
    // all others.
    
    for(int target: {1, 0}) {
        vector<vector<int>> mat(n, vector<int>(n, 0));
        for(int i = 0; i < n; i++) {
            for(int c: groups[i]) {
                mat[c][i] = 1;
            }
        }

        GaussBitset<200> g(n);
        for(int c = 0; c < n; c++) {
            g.add_equation(mat[c], initial[c] ^ target);
        }

        g.eliminate();
        if(!g.has_solution()) {
            continue;
        }
        auto sol = g.any_solution();
        vector<int> clicks;
        for(int i = 0; i < n; i++) {
            if(sol[i]) {
                clicks.push_back(i + 1);
            }
        }
        cout << clicks.size() << '\n';
        for(size_t i = 0; i < clicks.size(); i++) {
            cout << clicks[i] << " \n"[i + 1 == clicks.size()];
        }
        if(clicks.empty()) {
            cout << '\n';
        }
        return;
    }
    cout << -1 << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

4. Python Solution

```python
import sys

# Solve linear system over GF(2) using integer bitmasks.
# Since N <= 200, we can store each row as an int with (N+1) bits:
# bits [0..N-1] are coefficients, bit N is RHS.
def gauss_xor(rows, n):
    """
    rows: list[int], each is bitmask of length n+1 (coeffs + rhs at bit n)
    n: number of variables
    returns: (solvable: bool, where: list[int], rows: list[int])
             where[col] = pivot row index or -1 if free
             rows is transformed in-place to reduced form
    """
    where = [-1] * n
    r = 0  # current pivot row
    m = len(rows)

    for c in range(n):
        if r >= m:
            break

        # Find pivot row with coefficient 1 in column c
        pivot = -1
        mask = 1 << c
        for i in range(r, m):
            if rows[i] & mask:
                pivot = i
                break
        if pivot == -1:
            continue  # free variable

        # Swap pivot into position r
        rows[r], rows[pivot] = rows[pivot], rows[r]
        where[c] = r

        # Eliminate column c from all other rows (Gauss-Jordan)
        for i in range(m):
            if i != r and (rows[i] & mask):
                rows[i] ^= rows[r]

        r += 1

    # Check inconsistency: 0 coefficients but RHS = 1
    rhs_bit = 1 << n
    for i in range(r, m):
        coeffs = rows[i] & (rhs_bit - 1)  # lower n bits
        rhs = (rows[i] >> n) & 1
        if coeffs == 0 and rhs == 1:
            return False, where, rows

    return True, where, rows


def any_solution(where, rows, n):
    """
    With Gauss-Jordan form and free variables = 0, pivot variable equals RHS of its row.
    """
    x = [0] * n
    for var in range(n):
        pr = where[var]
        if pr != -1:
            x[var] = (rows[pr] >> n) & 1
    return x


def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n = int(next(it))

    groups = []
    for _ in range(n):
        k = int(next(it))
        cur = []
        for _ in range(k):
            cur.append(int(next(it)) - 1)  # to 0-based
        groups.append(cur)

    initial = [int(next(it)) for _ in range(n)]

    # Build coefficient matrix in bit form: equation per cell c:
    # XOR of x[i] where c in S_i equals initial[c] XOR target
    #
    # We'll first build for each cell c a bitmask of which variables affect it.
    coeff_mask = [0] * n
    for i in range(n):
        for c in groups[i]:
            coeff_mask[c] |= (1 << i)

    for target in (1, 0):
        rows = []
        for c in range(n):
            rhs = initial[c] ^ target
            row = coeff_mask[c] | (rhs << n)
            rows.append(row)

        solvable, where, reduced = gauss_xor(rows, n)
        if not solvable:
            continue

        x = any_solution(where, reduced, n)
        clicks = [str(i + 1) for i, v in enumerate(x) if v == 1]

        sys.stdout.write(str(len(clicks)) + "\n")
        if clicks:
            sys.stdout.write(" ".join(clicks) + "\n")
        else:
            # match the C++ behavior: extra blank line after 0
            sys.stdout.write("\n")
        return

    sys.stdout.write("-1\n")


if __name__ == "__main__":
    main()
```

5. Compressed Editorial

Model each click i as a binary variable x_i (clicked or not). Clicking toggles colors, so final color of cell c is a_c ⊕ (XOR of x_i over i with c ∈ S_i). We need all final colors equal to a target t ∈ {0,1}, giving N linear equations over GF(2): XOR of x_i over i with c ∈ S_i = a_c ⊕ t. Try t=1 and t=0. For each, solve Mx=b via Gaussian elimination mod 2 (bitset/bitmask optimization for speed). If consistent, set free vars to 0 and read pivot vars from RHS to obtain a valid solution; output indices with x_i=1. If neither target is solvable, print -1.
