1) Abridged problem statement
- You are given N unique team names and two full rankings (permutations) of these teams.
- The final ranking must satisfy: for every M where the sets of the first M teams in both rankings are identical, the first M in the final ranking is exactly that set.
- Whenever the rule does not fix the relative order of some teams, they must be ordered alphabetically.
- Output the final ranking (one team per line).

2) Detailed editorial
Key idea
- The rule fixes “cut positions” M where the set of the first M teams is the same in both rankings. Between two consecutive cuts, nothing else is fixed, so teams in that block must be alphabetically sorted.
- There are two natural ways to compute the final order:
  A) Graph/SCC method (the provided C++ solution)
  B) A direct linear scan using positions and block boundaries (simpler)

A) Graph/SCC method (as in the provided C++ code)
- Build a directed graph on the N teams (we map each team to its index in the first ranking).
- Add edges that go from each item to the previous item (i -> i-1) in:
  - ranking 1, and
  - ranking 2 (converted to indices of ranking 1 via name_to_id).
- Intuition: In each ranking, there is a directed path from a later team to any earlier team. By adding edges for both rankings, whenever two teams must be comparable due to the rule, you’ll have one-way reachability; when two teams are mutually reachable (a cycle), their relative order cannot be deduced and they must be alphabetically ordered.
- Compute strongly connected components (SCCs). Teams in the same SCC form a “tie block”.
- The condensation of this graph (SCC DAG) is a chain: because we added i -> i-1 for the first ranking, there’s always a path from any later index to any earlier index, so any two SCCs are comparable.
- Order the SCCs by the minimum index (in ranking 1) they contain. This is a topological order of the chain.
- Within each SCC, sort team names alphabetically (the code implicitly does this by sorting pairs (min_index_of_component, name)).
- Output teams in the order of components; teams within the same component in lexicographical order.

Correctness
- If two teams lie in different SCCs, then in the condensation chain one component precedes the other. Hence the relative order between those components is fixed and respects both rankings’ prefix-equality rule.
- If two teams lie in the same SCC, there is a cycle between them using edges that come only from adjacent constraints in both rankings. That means no M exists where exactly one of them is in the prefix set in both rankings. So their relative order is not determined by the rule, hence we sort them alphabetically.

Complexity
- Building the graph uses O(N) edges (2N-2).
- Kosaraju’s SCC is O(N).
- Sorting by groups and lexicographically is O(N log N).
- Memory: O(N).

B) Simpler alternative (linear scan + block sorting)
- Map each team to its position in the second ranking: pos2[name].
- Iterate the first ranking from left to right; maintain max_pos = max(pos2[a[i]] over the current prefix).
- Whenever max_pos == i, the set of the first i+1 in ranking 1 matches the set of the first i+1 in ranking 2; that index is a block boundary.
- Alphabetically sort the teams of the current block and append to the answer; start a new block after it.
- This yields the same result and is very simple. Complexity is also O(N log N) due to sorting within blocks.

3) C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/graph/scc.hpp>

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;
};

class StronglyConnectedComponents {
  private:
    vector<bool> visited;

    void dfs1(int u) {
        visited[u] = true;
        for(int v: adj[u]) {
            if(!visited[v]) {
                dfs1(v);
            }
        }

        top_sort.push_back(u);
    }

    void dfs2(int u) {
        for(int v: radj[u]) {
            if(comp[v] == -1) {
                comp[v] = comp[u];
                dfs2(v);
            }
        }
    }

  public:
    int n;
    vector<vector<int>> adj, radj;
    vector<int> comp, comp_ids, top_sort;

    StronglyConnectedComponents() {}
    StronglyConnectedComponents(int _n) { init(_n); }

    void add_edge(int u, int v) {
        adj[u].push_back(v);
        radj[v].push_back(u);
    }

    void init(int _n) {
        n = _n;
        comp_ids.clear();
        top_sort.clear();
        adj.assign(n, {});
        radj.assign(n, {});
    }

    void find_components() {
        comp.assign(n, -1);
        visited.assign(n, false);

        for(int i = 0; i < n; i++) {
            if(!visited[i]) {
                dfs1(i);
            }
        }

        reverse(top_sort.begin(), top_sort.end());
        for(int u: top_sort) {
            if(comp[u] == -1) {
                comp[u] = (int)comp_ids.size();
                comp_ids.push_back(comp[u]);
                dfs2(u);
            }
        }
    }
};

int n;
vector<string> ord1, ord2;

void read() {
    cin >> n;
    ord1.resize(n);
    ord2.resize(n);
    cin >> ord1 >> ord2;
}

void solve() {
    // The solution here is fairly simple - we build a graph out of the
    // two orders, and notice that if there are cycles we can't deduce
    // the relative order of the elements of said cycle (and so we
    // revert to lexicographical). This should immediately lead us to think
    // about strongly connected components. Because the edges are simply given
    // by two orders, we can notice that after compressing into strongly
    // connected components we are left with a chain. Part of the problem
    // statement talks about having the same "set" in the answer for every
    // prefix if the two orders match in that prefix. We can notice that the
    // above solution will satisfy this trivially, as if the prefix match, so do
    // the suffixes, so we can cut the 2 edges coming from the two orders and
    // get two disjoint graphs.

    map<string, int> name_to_id;
    for(int i = 0; i < n; i++) {
        name_to_id[ord1[i]] = i;
    }

    StronglyConnectedComponents scc(n);
    for(int i = 1; i < n; i++) {
        scc.add_edge(i, i - 1);
        scc.add_edge(name_to_id[ord2[i]], name_to_id[ord2[i - 1]]);
    }

    scc.find_components();

    vector<int> min_id_in_comp(n, n);
    for(int i = 0; i < n; i++) {
        min_id_in_comp[scc.comp[i]] = min(min_id_in_comp[scc.comp[i]], i);
    }

    vector<pair<int, string>> result;
    for(int i = 0; i < n; i++) {
        result.push_back({min_id_in_comp[scc.comp[i]], ord1[i]});
    }

    sort(result.begin(), result.end());
    for(int i = 0; i < n; i++) {
        cout << result[i].second << 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 (detailed comments, direct linear scan + block sort)
Note: This is the simpler alternative described in the editorial (method B). It produces exactly the same final ranking required by the statement.

```python
import sys

def main():
    data = sys.stdin.read().strip().split()
    it = iter(data)

    # Read number of teams
    n = int(next(it))

    # Read the two rankings
    ord1 = [next(it) for _ in range(n)]
    ord2 = [next(it) for _ in range(n)]

    # Map each team to its position in the second ranking (0-based)
    pos2 = {name: i for i, name in enumerate(ord2)}

    # We will scan ord1 left-to-right and maintain the maximum pos2 seen so far.
    # When max_pos2 == i, the set of ord1[0..i] equals the set of ord2[0..i].
    # That means we have a "block boundary". We sort that block alphabetically.
    answer = []            # Collect the final order here
    block = []             # Current block of names since last boundary
    max_pos2 = -1          # Max position in ord2 among names in current block

    for i, name in enumerate(ord1):
        block.append(name)                 # Include this name in current block
        max_pos2 = max(max_pos2, pos2[name])  # Update the furthest ord2 position reached

        # If the prefix sets match in both orders at this i,
        # we flush the block sorted alphabetically.
        if max_pos2 == i:
            block.sort()                   # Relative order not fixed -> alphabetical
            answer.extend(block)           # Add to final answer
            block.clear()                  # Start a new block for the next segment

    # Safety: in well-formed inputs, block must be empty here.
    # But if not, sort and append (handles any trailing block).
    if block:
        block.sort()
        answer.extend(block)

    # Output the final ranking
    sys.stdout.write("\n".join(answer))

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

Why this works
- At step i, max_pos2 equals the highest index in ord2 among teams seen so far in ord1. If max_pos2 == i, we have encountered exactly the set of teams placed in the first i+1 positions of ord2, hence the prefix sets coincide. Therefore, the current block’s membership is fixed, but not its internal order, so we sort it alphabetically.
- This exactly implements the rule and yields the same result as the SCC approach.

Complexity
- O(n) scanning plus sorting inside blocks; worst-case O(n log n).

5) Compressed editorial
- Build precedence constraints from adjacent pairs in both rankings by adding edges from each team to its predecessor in each ranking.
- Teams that mutually reach each other (SCCs) are exactly those whose relative order cannot be deduced from the “equal prefix sets” rule; they must be sorted alphabetically.
- The SCC condensation is a chain since we always add edges i -> i-1 for ranking 1, so components can be ordered by the smallest index they contain in ranking 1.
- Output teams sorted by (component_min_index, name).
- Alternatively, scan ranking 1, track the maximum position reached in ranking 2; whenever it equals the current index, you’ve found a block boundary. Sort each block alphabetically and concatenate.