## 1) Concise abridged problem statement

There are \(N\) cells ( \(1 \le N \le 200\) ), 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 (solution explanation)

### 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:
- \(x_i \in \{0,1\}\) = whether we click cell \(i\).
- \(a_j \in \{0,1\}\) = 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:
\[
a_j \oplus \bigoplus_{i: j \in S_i} x_i
\]
(where \(\oplus\) is XOR).

We want all final colors equal to some target \(t \in \{0,1\}\):
\[
a_j \oplus \bigoplus_{i: j \in S_i} x_i = t \quad \forall j
\]
Rearrange:
\[
\bigoplus_{i: j \in S_i} x_i = a_j \oplus t
\]

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:
- \(t=0\) (make everything white)
- \(t=1\) (make everything black)

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

### Building the matrix

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

### Gaussian elimination over GF(2) with bitsets

\(N \le 200\). 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`.

---

## 3) Provided C++ solution with detailed line-by-line comments

```cpp
#include <bits/stdc++.h>              // Includes almost all standard headers (GCC extension)

using namespace std;

// Pretty-print a pair as "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

// Read all elements of a vector<T> from input
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {                // iterate by reference so we can assign into vector
        in >> x;
    }
    return in;
};

// Print all elements of a vector<T>, each followed by a space
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// Gaussian elimination over GF(2) using bitsets.
// N is a compile-time maximum number of variables (here used as 200).
template<size_t N>
class GaussBitset {
  public:
    int n_var;                        // actual number of variables (<= N)
    vector<bitset<N + 1>> rows;       // each row: N coeff bits + 1 RHS bit at index n_var
    vector<int> where;                // where[col] = row index used as pivot for that variable, or -1
    bool solvable = false;            // becomes true after elimination if system is consistent

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

    // Add one equation: sum(coeffs[i]*x_i) = rhs (mod 2)
    void add_equation(const vector<int>& coeffs, int rhs) {
        bitset<N + 1> row;            // bitset initialized to all zeros
        for(int i = 0; i < n_var; i++) {
            if(coeffs[i] % 2) {       // in GF(2), we only care about parity
                row[i] = 1;           // set coefficient bit
            }
        }
        row[n_var] = rhs % 2;         // RHS stored in the last used position
        rows.push_back(row);          // append to matrix
    }

    // Perform Gauss-Jordan elimination over GF(2)
    void eliminate() {
        int n_eq = (int)rows.size();  // number of equations
        where.assign(n_var, -1);      // initially no pivot for any variable
        int row = 0;                  // current pivot row

        // iterate over columns (variables)
        for(int col = 0; col < n_var && row < n_eq; col++) {
            int sel = -1;             // row selected as pivot
            // find a row >= current pivot row that has 1 in this column
            for(int i = row; i < n_eq; i++) {
                if(rows[i][col]) {
                    sel = i;
                    break;
                }
            }
            if(sel == -1) {           // no pivot in this column => free variable
                continue;
            }

            swap(rows[sel], rows[row]);   // move pivot row into position
            where[col] = row;             // record pivot location for this variable

            // eliminate this column from all other rows (Gauss-Jordan style)
            for(int i = 0; i < n_eq; i++) {
                if(i != row && rows[i][col]) {
                    rows[i] ^= rows[row]; // XOR rows => subtract in GF(2)
                }
            }
            row++;                        // advance to next pivot row
        }

        // After elimination, check for inconsistency: 0 = 1 rows
        solvable = true;
        for(int i = row; i < n_eq; i++) { // rows without pivots
            if(rows[i][n_var]) {          // RHS is 1 but LHS is all zeros
                solvable = false;
                break;
            }
        }
    }

    bool has_solution() const { return solvable; }

    // Return indices of variables that are free (not used as pivots)
    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;
    }

    // Produce one solution by setting all free vars = 0 and reading pivot vars from RHS.
    vector<int> any_solution() const {
        vector<int> x(n_var, 0);          // default all zeros
        for(int v = 0; v < n_var; v++) {
            if(where[v] != -1) {          // pivot variable
                x[v] = rows[where[v]][n_var]; // equals RHS when free vars are 0
            }
        }
        return x;
    }
};

int n;                                   // number of cells/variables
vector<vector<int>> groups;              // groups[i] = list of cells toggled by clicking i
vector<int> initial;                     // initial colors

void read() {
    cin >> n;
    groups.assign(n, {});
    for(int i = 0; i < n; i++) {
        int k;
        cin >> k;                        // size of set for click i
        groups[i].resize(k);
        for(auto& x: groups[i]) {
            cin >> x;                    // read 1-based cell index
            x--;                         // convert to 0-based
        }
    }
    initial.assign(n, 0);
    for(auto& x: initial) {
        cin >> x;                        // read initial colors (0/1)
    }
}

void solve() {
    // We try both targets: make all cells color 1 (black), or all 0 (white).
    // Build a linear system over GF(2):
    //   For each cell c: XOR of x[i] for which clicking i toggles c  = initial[c] XOR target.

    for(int target: {1, 0}) {            // attempt target = 1 first, then 0
        // Build coefficient matrix mat[c][i] = 1 iff clicking i affects cell c.
        vector<vector<int>> mat(n, vector<int>(n, 0));
        for(int i = 0; i < n; i++) {     // for each click i
            for(int c: groups[i]) {      // for each cell c toggled by click i
                mat[c][i] = 1;           // coefficient of variable x_i in equation for cell c
            }
        }

        GaussBitset<200> g(n);           // solver with max size 200, actual vars n

        // Add N equations, one per cell c
        for(int c = 0; c < n; c++) {
            // RHS is initial[c] XOR target (since we want final[c] == target)
            g.add_equation(mat[c], initial[c] ^ target);
        }

        g.eliminate();                   // perform elimination
        if(!g.has_solution()) {          // no solution for this target color
            continue;
        }

        auto sol = g.any_solution();     // extract some solution
        vector<int> clicks;
        for(int i = 0; i < n; i++) {
            if(sol[i]) {                 // if x_i = 1, click i
                clicks.push_back(i + 1); // convert back to 1-based
            }
        }

        // Output solution
        cout << clicks.size() << '\n';
        for(size_t i = 0; i < clicks.size(); i++) {
            cout << clicks[i] << " \n"[i + 1 == clicks.size()]; // space or newline at end
        }
        if(clicks.empty()) {             // matches expected formatting (blank line after 0)
            cout << '\n';
        }
        return;                          // stop after finding any valid target
    }

    // Neither all-1 nor all-0 reachable
    cout << -1 << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);    // fast I/O
    cin.tie(nullptr);

    int T = 1;                           // single test case
    // cin >> T;                         // (not used)
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```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:
\[
a_c \oplus \bigoplus_{i: c \in S_i} x_i
\]
We need all final colors equal to a target \(t\in\{0,1\}\), giving \(N\) linear equations over GF(2):
\[
\bigoplus_{i: c \in S_i} x_i = a_c \oplus t
\]
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`.