1. Abridged Problem Statement  
Given an undirected social graph of N users (numbered 1…N) and a specific user x, find all users c such that:  
- c ≠ 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.

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(N²) 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  
• Initialize an empty list result.  
• 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 result.  

Step 3 – Output  
• Sort result (we can build it in increasing order anyway since we scan u from 1..N).  
• Print result.size(), then the elements of result on one line.

Time Complexity: O(N²). Memory: O(N²) for the adjacency matrix.

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Maximum number of users is 50, so 1<<6 = 64 is safe for an adjacency matrix
const int MAXN = (1 << 6);

int n, x;                 // n = number of users, x = target user
int adj[MAXN][MAXN];      // adjacency matrix: adj[i][j] = 1 if i and j are friends

// Read input and build the adjacency matrix
void read() {
    cin >> n >> x;
    // For each user i, read their friend count and friend list
    for (int i = 1; i <= n; i++) {
        int cnt;
        cin >> cnt;
        while (cnt--) {
            int f;
            cin >> f;
            // Mark friendship in both directions
            adj[i][f] = 1;
            adj[f][i] = 1;
        }
    }
}

// Find and print friends-of-friends of user x
void solve() {
    vector<int> result;

    // Consider every user i from 1 to n
    for (int i = 1; i <= n; i++) {
        // Skip if same as x or already a direct friend of x
        if (i == x || adj[x][i] == 1) continue;

        // Check if i is connected to x via some mutual friend
        bool isFoF = false;
        for (int f = 1; f <= n; f++) {
            // f must be a direct friend of x and also a friend of i
            if (adj[x][f] && adj[f][i]) {
                isFoF = true;
                break;
            }
        }

        // If found, record i
        if (isFoF) {
            result.push_back(i);
        }
    }

    // Output the count
    cout << result.size() << '\n';
    // Output the list in increasing order (already in order since i goes 1..n)
    for (int u : result) {
        cout << u << " ";
    }
    cout << '\n';
}

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

    read();    // Build the graph
    solve();   // Compute and print friends of friends
    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×N adjacency matrix (or adjacency sets).  
• For each user u ≠ x and not a direct friend of x, check if ∃ f ∈ friends(x) ∩ friends(u).  
• Collect such u, sort them, and output count + list.  
Complexity: O(N²), N ≤ 50.