## 1. Abridged problem statement

Given an undirected social graph of N users (numbered 1 to N) and a specific user x, find all users c such that:
- c is not x,
- c is *not* a direct friend of x,
- there exists some b who *is* a friend of x and also a friend of c.

Output the count of such "friends of friends" of x, then list them in increasing order.

Constraints: N <= 50.

---

## 2. Detailed editorial

We want all nodes at graph-distance exactly 2 from x, excluding x itself and x's direct neighbors. N is at most 50, so an O(N^2) solution is trivial.

**Step 1 - Read and store the graph.**
Use an adjacency matrix `adj[i][j]` (size (N+1) by (N+1)). For each user i, read their friend list and set `adj[i][friend] = adj[friend][i] = 1`.

**Step 2 - Identify friends of friends.**
For each candidate user u from 1 to N:
- Skip if u == x (cannot be x itself).
- Skip if `adj[x][u] == 1` (u is a direct friend of x).
- Otherwise, check if there exists some friend f of x such that `adj[f][u] == 1`. If yes, add u to the result.

**Step 3 - Output.**
The result is already in increasing order since we scan u from 1 to N. Print the count, then the elements.

Time Complexity: O(N^2). Memory: O(N^2) for the adjacency matrix.

---

## 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, x;
vector<vector<int>> adj;

void read() {
    cin >> n >> x;
    adj.assign(n + 1, vector<int>(n + 1, 0));
    for(int i = 1; i <= n; i++) {
        int cnt;
        cin >> cnt;
        while(cnt--) {
            int f;
            cin >> f;
            adj[f][i] = 1;
            adj[i][f] = 1;
        }
    }
}

void solve() {
    // A friend of a friend of x is any user i that is not x, is not a direct
    // friend of x, yet shares a common friend o with x (o is a friend of x and
    // o is a friend of i). Scan every candidate i and, using the adjacency
    // matrix, look for such an intermediate o; emit the matching users in
    // increasing order.

    vector<int> li;
    for(int i = 1; i <= n; i++) {
        if(i == x || adj[i][x]) {
            continue;
        }

        bool ok = false;
        for(int o = 1; o <= n; o++) {
            if(adj[x][o] && adj[o][i]) {
                ok = true;
            }
        }

        if(ok) {
            li.push_back(i);
        }
    }

    cout << li.size() << '\n';
    cout << li << '\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();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 4. Python solution with detailed comments

```python
def main():
    import sys
    data = sys.stdin.read().split()
    it = iter(data)

    # Read number of users n and target user x
    n = int(next(it))
    x = int(next(it))

    # Build adjacency sets for each user
    # friends[i] is a set of direct friends of i
    friends = [set() for _ in range(n+1)]
    for i in range(1, n+1):
        d = int(next(it))
        for _ in range(d):
            f = int(next(it))
            friends[i].add(f)
            friends[f].add(i)

    result = []
    # Examine each candidate user u
    for u in range(1, n+1):
        if u == x:
            continue                  # skip x itself
        if u in friends[x]:
            continue                  # skip direct friends of x

        # Check if u shares any mutual friend with x
        # i.e., intersection of friends[x] and friends[u] is non-empty
        if friends[x].intersection(friends[u]):
            result.append(u)

    # Output count and sorted list
    result.sort()
    print(len(result))
    if result:
        print(*result)

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

---

## 5. Compressed editorial

Build an N x N adjacency matrix (or adjacency sets). For each user u that is not x and not a direct friend of x, check if there exists f in friends(x) that is also a friend of u. Collect such u, sort them, and output count + list. Complexity: O(N^2), N <= 50.
