1. **Abridged problem statement**

- There are \(N\) cities, \(M\) airline companies.
- Between every pair of cities \(i \neq j\) there is exactly one direct flight, owned by exactly one company (numbered \(1..M\)).
- The president wants to choose some companies; all their flights become free.
- Using only free flights, any city must be reachable from any other in at most 3 flights (i.e. path length \(\le 3\)).
- Due to antitrust rules, he may buy **at most \(\lfloor (M+1)/2 \rfloor\)** companies.
- Output:
  - If possible: how many and which company indices to buy (any valid set).
  - If impossible: output `-1`.

---

2. **Detailed editorial**

### 2.1. Reformulation as graph problem

- Cities: vertices \(1..N\).
- Each unordered pair \(\{i, j\}\) has exactly one edge labeled by a company color \(c(i,j) \in \{1..M\}\).
- If we choose a subset \(S \subseteq \{1..M\}\), all edges whose color is in \(S\) are “free”.
- Condition: In the subgraph induced by free edges, the shortest path between every pair of vertices has length \(\le 3\).
- Restriction: \(|S| \le \lfloor (M+1)/2 \rfloor\).

We must find any such \(S\), or say that none exists.

### 2.2. Key structural insight: 2-color case is always solvable

First, shrink the problem to **2 colors** (say black and white). Assume:

- Every edge is either black or white.
- We want to find a color such that, in the subgraph containing only edges of that color, the distance between any two vertices is at most 3.

**Claim (crucial lemma).** For any complete graph on \(N \ge 1\) vertices whose edges are colored black/white, **at least one color induces a graph where every pair of vertices has graph distance \(\le 3\)**.

Equivalently: you cannot have both:
- There exist vertices \(u_b, v_b\) such that their shortest black-only path is > 3.
- There exist vertices \(u_w, v_w\) such that their shortest white-only path is > 3.

Sketch of reasoning (as in the code comments):

1. For \(N = 4\):
   - Consider an arbitrary 2-coloring of all \(\binom{4}{2} = 6\) edges.
   - It's known (and can be checked exhaustively, or via the complement argument below) that at least one of the two color graphs has diameter \(\le 3\).
   - Complement argument: For any (uncolored) simple graph \(H\) on 4 vertices, either \(H\) or its complement \(\overline{H}\) is connected, and in any connected graph on 4 vertices, shortest-path distance between any two vertices is at most 3 (because any simple path can use at most 3 edges).

2. For general \(N\):
   - Suppose for contradiction that **both** monochromatic graphs (black and white) contain at least one pair of vertices at distance > 3.
   - Let \(u_b, v_b\) be a pair at distance > 3 in black.
   - Let \(u_w, v_w\) be a pair at distance > 3 in white.
   - Consider the induced subgraph on the 4 vertices \(\{u_b, v_b, u_w, v_w\}\). (Some of these may coincide but then distances can only get smaller, so we can treat the problematic pairs as distinct.)
   - On these 4 vertices, restrict to black edges only; the distance between \(u_b\) and \(v_b\) is still > 3.
   - Similarly, in white-only edges, the distance between \(u_w\) and \(v_w\) is still > 3.
   - But this contradicts the \(N = 4\) lemma: on 4 vertices, it is impossible that **both** colors have diameter > 3.
   - Thus the assumption is false, and at least one color must yield diameter \(\le 3\).

So, with 2 colors, we are assured that there exists at least one “good” color whose subgraph has all pairwise distances \(\le 3\).

### 2.3. Reducing M colors to 2 colors via parity

We have \(M\) colors. The constraint on number of companies:

\[
|S| \le \left\lfloor\frac{M + 1}{2}\right\rfloor.
\]

Observe:
- Number of odd indices in \(1..M\) is \(\lceil M/2 \rceil\).
- Number of even indices in \(1..M\) is \(\lfloor M/2 \rfloor\).
- \(\lfloor(M+1)/2\rfloor = \lceil M/2 \rceil\).
  - For odd \(M = 2t+1\): both sides = \(t+1\).
  - For even \(M = 2t\): both sides = \(t\).

So:
- You can choose at most \(\lfloor(M+1)/2\rfloor\) companies.
- That number equals the size of the larger parity class (odd or even).
- Therefore, either all odd colors or all even colors **fit** within the limit.

We now **map the multi-color problem to a 2-color problem**:

- For each edge color \(c\):
  - Replace it by \(c \bmod 2\):
    - Parity 1 → “odd color” (we’ll treat as black, say).
    - Parity 0 → “even color” (white).
- Now each edge has one of 2 parities: 0 or 1.

By the lemma, among parity 0 and parity 1, **at least one parity’s subgraph has diameter \(\le 3\)**.

We need to find which parity works.

### 2.4. Checking a parity with Floyd–Warshall

Let’s fix the representation:

- After `G[i][j] %= 2`, we have an \(N \times N\) matrix where:
  - `G[i][j] == 0` means “edge \(i-j\) has even company index”.
  - `G[i][j] == 1` means “edge \(i-j\) has odd company index”.

We’d like a function that checks if *even* edges alone provide diameter \(\le 3\). That is exactly `solve_even()`:

1. Build `dist` matrix (`n x n`), initialized to a large “infinite” value, say `1e9`.
2. Set `dist[i][i] = 0` for all `i`.
3. For each pair `(i, j)` with `i < j`:
   - If `G[i][j] == 0` (even parity):
     - Set `dist[i][j] = dist[j][i] = 1` (direct free edge).
4. Run Floyd–Warshall:
   - For all `k, i, j`, relax:
     - `dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])`.
5. Finally, check that for all `i, j`, `dist[i][j] <= 3`. If any is > 3, then even parity **fails** (not good enough), return false. Otherwise return true.

So:

- If `solve_even()` returns `true`, then using all **even** companies satisfies diameter ≤ 3.
- If `solve_even()` returns `false`, then even companies alone are not enough; but by the lemma, this implies that **odd** companies alone must work.

### 2.5. Choosing actual company indices and respecting the limit

We must output actual company numbers (1..M) to buy.

Strategy used in the solution:

```cpp
vector<int> ans;
for(int i = 1 + solve_even(); i <= k; i += 2) {
    ans.push_back(i);
}
```

Analyze:

- If `solve_even()` is `true`:
  - `1 + solve_even()` = 2.
  - Loop: `i = 2, 4, 6, ..., ≤ M`.
  - So we choose all **even** companies.
- If `solve_even()` is `false`:
  - `1 + solve_even()` = 1.
  - Loop: `i = 1, 3, 5, ..., ≤ M`.
  - So we choose all **odd** companies.

Why is that always within the allowed limit?

- If `solve_even()` is true, then the number of even indices is `floor(M/2) ≤ floor((M+1)/2)`, OK.
- If `solve_even()` is false, then by lemma, odds must work; their count is `ceil(M/2) = floor((M+1)/2)`, exactly the maximum allowed.

Thus the solution simply:
1. Reduces all colors to parity (`G[i][j] %= 2`).
2. Uses Floyd–Warshall to test whether parity 0 (even indices) suffices.
3. If yes, chooses the even companies; if not, chooses the odd ones.
4. Prints the count and list.

Crucially, the lower-level 2-color lemma guarantees that **one of the two parities always works**, so we never print `-1`.

Complexity:
- `N ≤ 200`.
- Floyd–Warshall: \(O(N^3) = 8\cdot 10^6\) operations, easily fast enough.
- Memory: `O(N^2)` = 40,000 ints, well within limits.

---

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;
};

int n, k;
vector<vector<int>> G;

void read() {
    cin >> n >> k;
    G.assign(n, vector<int>(n, 0));
    cin >> G;
}

bool solve_even() {
    vector<vector<int>> dist(n, vector<int>(n, (int)1e9));
    for(int i = 0; i < n; i++) {
        dist[i][i] = 0;
    }
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            if(G[i][j] == 0) {
                dist[i][j] = 1;
                dist[j][i] = 1;
            }
        }
    }

    for(int k = 0; k < n; k++) {
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
            }
        }
    }

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(dist[i][j] > 3) {
                return false;
            }
        }
    }
    return true;
}

void solve() {
    // In problems like this, it's always good to ask the question of when this
    // is actually possible. Turns out it always is. In particular, let's
    // consider the case of only 2 colours, as reducing k colours to that case
    // is trivial - just split into "odd" and "even" colours.
    //
    // The claim is that for 2 colours, one of them always satisfies that the
    // distance between each pair is <= 3. Let's consider the case of n=4 first.
    // We can go through all examples to convince ourselves, but there is a
    // simpler argument for why this is true - for any graph G, either G or K_n
    // - G is connected (this is a standard results which we can get convinced
    // about by thinking about the complementary edges and that they connect all
    // connected components in G), and with n nodes the distance is always <= 3
    // (as 4 edges means a cycle and so not a simple path).
    //
    // Now let's look at n > 4 and assume for contradiction. This means that for
    // both colours, there are 2 vertices (u_black, v_black) and (u_white,
    // v_white) such that the black distance between u_black and v_black is > 3,
    // and the white distance between u_white and v_white is > 3. However, we
    // already have a result for n = 4: consider the subgraph with V = {u_black,
    // v_black, u_white, v_black} and the result implying that there is at least
    // one colour that has distance less than or equal to 3 between all pairs.
    // This means there is a trivial contradiction between one of the two
    // assumptions.
    //
    // Now that we have some results, let's actually solve the problem. The
    // first step is to split the K colours into 2. This will be done via the
    // parity. Afterwards, we can just do Floyd to check if all distances are
    // less than or equal to 3, as the constraints are n <= 200. We should
    // technically be able to do this a bit faster with bitsets - it's enought
    // to compute G^1, G^2 and G^3, but for this problem Floyd is enough.

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            G[i][j] = G[i][j] % 2;
        }
    }

    vector<int> ans;
    for(int i = 1 + solve_even(); i <= k; i += 2) {
        ans.push_back(i);
    }

    cout << ans.size() << endl;
    cout << ans << endl;
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

4. **Python solution with detailed comments**

```python
import sys

# We will implement the same logic as the C++ code, but in Python.
# Steps:
# 1) Read N (cities) and M (companies).
# 2) Read the matrix of company indices.
# 3) Reduce all entries modulo 2 (parity: 0 = even, 1 = odd).
# 4) Run Floyd–Warshall using only even edges to test if "even" parity works.
# 5) If it works, output all even company indices, else all odd indices.

def read_ints():
    """Read a line and parse it into a list of ints."""
    return list(map(int, sys.stdin.readline().split()))

def solve():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)

    # Read N and M
    n = int(next(it))  # number of cities
    m = int(next(it))  # number of companies

    # Read the N x N matrix G of company indices
    # G[i][j] is in 0..m, with 0 on diagonal
    G = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(n):
            G[i][j] = int(next(it))

    # Reduce edges to parity: 0 => even company index, 1 => odd index
    # We do this in-place to reuse G for parity matrix.
    for i in range(n):
        for j in range(n):
            G[i][j] %= 2

    # Function to check if using ONLY even edges (parity 0) gives
    # graph diameter <= 3 among all vertices.
    def solve_even():
        INF = 10**9
        # Initialize distance matrix
        dist = [[INF] * n for _ in range(n)]

        # Distance from node to itself is 0
        for i in range(n):
            dist[i][i] = 0

        # For edges with even company index (parity 0),
        # set distance between the endpoints to 1.
        for i in range(n):
            for j in range(i + 1, n):
                if G[i][j] == 0:  # even edge
                    dist[i][j] = 1
                    dist[j][i] = 1

        # Floyd–Warshall: compute all-pairs shortest paths
        for k in range(n):
            # Standard triple loop: i, j, k
            # Using local variables for small speed-up in Python.
            dk = dist[k]
            for i in range(n):
                di = dist[i]
                ik = di[k]
                if ik == INF:
                    # If i cannot reach k, no need to use k as intermediate
                    continue
                for j in range(n):
                    # Try to relax dist[i][j] via k
                    alt = ik + dk[j]
                    if alt < di[j]:
                        di[j] = alt

        # Check if any pair exceeds distance 3
        for i in range(n):
            for j in range(n):
                if dist[i][j] > 3:
                    return False
        return True

    # Decide whether even parity works
    even_ok = solve_even()

    # Prepare list of companies to buy:
    # - If even_ok: buy all even companies (2,4,...)
    # - Otherwise: buy all odd companies (1,3,...)
    ans = []
    if even_ok:
        start = 2
    else:
        start = 1

    for company in range(start, m + 1, 2):
        ans.append(company)

    # Output answer.
    # Theoretically, the problem allows -1 if impossible,
    # but by the 2-color lemma, one parity always works, so we never output -1.
    out_lines = []
    out_lines.append(str(len(ans)))
    if ans:
        out_lines.append(" ".join(map(str, ans)))
    else:
        # Edge case: if m=0 (not possible per constraints, but to be safe)
        out_lines.append("")
    sys.stdout.write("\n".join(out_lines))


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

---

5. **Compressed editorial**

- Interpret flights as edges of a complete graph on \(N\) vertices, each edge colored by its owning company (1..M).
- Choosing some companies = taking all edges whose color is in a chosen set. We need this subgraph to have distance ≤ 3 between every pair of cities.
- Let’s reduce colors to **parity**:
  - Recolor each edge by \(c(i,j) \bmod 2\) → “even” vs “odd”.
  - We can buy either all even-index companies or all odd-index companies; each set’s size is at most \(\lfloor(M+1)/2\rfloor\), exactly the allowed limit.

Core lemma (2-color case):
- In any complete graph whose edges are colored with 2 colors, at least one color induces a graph of diameter ≤ 3.
- Proof sketch: For 4 vertices it's true by graph/complement argument. For general \(N\), if both colors had a pair at distance > 3, those 4 vertices (the two bad pairs) would contradict the \(N=4\) case.

Algorithm:
1. Read \(N, M\) and matrix of company indices.
2. Replace each company index by its parity (`%2`).
3. Test “even” edges:
   - Build an adjacency matrix with cost 1 for even edges, ∞ otherwise.
   - Run Floyd–Warshall to get all-pairs shortest paths.
   - If any distance > 3 → even fails, else even works.
4. If even works, output all even indices `2,4,...`; else output all odd indices `1,3,...`.

Complexity:
- Floyd–Warshall: \(O(N^3)\) with \(N ≤ 200\) → easily fits time.
- Memory: \(O(N^2)\).

Because one parity always works by the lemma, a solution always exists; we never print `-1`.