<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

281. Championship
Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



N teams participate in a championship. It consists of two contests; each of the teams participates in both. Each contest has its own rating. Then the final rating is made. If for some M (1≤ M≤ N) the sets of first M teams in both ratings are the same, then this set is the set of the first M teams in the final rating. Teams whose relative order cannot be determined using this rule are placed in alphabetical order.

You are given the contests' ratings. Your task is to generate the final rating.

Input
The first line of the input contains an integer N (1≤ N≤ 50000). The two ratings follow. Each of the ratings consists on N lines, each of which contains a team's name, being a string of at most 20 lowercase Latin letters and digits.

Output
Output the final rating in the same format as in the input.

Example(s)
sample input
sample output
6
spbifmo1
msu1
msu2
permsu
nsu1
nnsu
spbifmo1
msu1
permsu
msu2
nnsu
nsu1
spbifmo1
msu1
msu2
permsu
nnsu
nsu1



Novosibirsk SU Contest #2, by Novosibirsk Team #1

<|response|>
1) Abridged problem statement
- You are given N unique team names and two rankings (permutations) of these teams.
- The final ranking must satisfy: for any M where the sets of the first M teams in both rankings are identical, the first M teams in the final ranking are exactly that set (order still unspecified).
- Any relative order that is not fixed by this rule must be resolved by sorting teams alphabetically.
- Output the final ranking.

2) Key observations
- The rule only fixes which teams must appear among the first M positions whenever the two rankings have the same prefix set of size M. It does not fix the internal order within that set; hence those teams must be ordered alphabetically.
- “Cut positions” M are exactly positions where the set of the first M teams in ranking 1 equals the set of the first M teams in ranking 2.
- These cuts can be found by a single linear scan:
  - Map each team to its position in ranking 2: pos2[name].
  - Scan ranking 1 left to right and maintain max_pos = max(pos2[name] for teams seen so far).
  - Whenever max_pos equals the current index i, the first i+1 teams in both rankings have identical sets. This ends a block.
- Between two consecutive cuts, nothing else is determined; teams inside each block must be sorted alphabetically.

3) Full solution approach based on the observations
- Read N and the two rankings ord1 and ord2.
- Build pos2: a dictionary mapping each team name to its index in ord2.
- Initialize:
  - an empty list block to collect names since the previous cut,
  - an empty list answer for the final order,
  - max_pos2 = -1.
- For i from 0 to N-1:
  - Append ord1[i] to block.
  - Update max_pos2 = max(max_pos2, pos2[ord1[i]]).
  - If max_pos2 == i, we’ve found a cut:
    - sort block alphabetically,
    - append the sorted block to answer,
    - clear block and continue scanning.
- Output the names in answer, one per line.
- Correctness:
  - At each cut i, the set of ord1[0..i] equals the set of ord2[0..i], so the block’s membership is fixed by the rule. Nothing else constrains their internal order, so sorting alphabetically is required.
  - Concatenating the sorted blocks yields a ranking that satisfies all prefix-set constraints.
- Complexity:
  - O(N) to scan and track max_pos2.
  - O(sum over blocks of k log k) for sorting inside blocks; worst-case O(N log N).
  - Memory O(N).

4) 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;
}
```

5) Python implementation with detailed comments
```python
import sys

def main():
    data = sys.stdin.buffer.read().split()
    if not data:
        return

    n = int(data[0])

    # Keep names as bytes for speed; lexicographic order on bytes matches ASCII
    # and thus works for lowercase letters and digits.
    ord1 = data[1 : 1 + n]
    ord2 = data[1 + n : 1 + 2 * n]

    # Map team -> position in ord2
    pos2 = {name: i for i, name in enumerate(ord2)}

    answer = []
    block = []
    max_pos2 = -1

    # Scan ord1; find cut positions where prefix sets match.
    for i, name in enumerate(ord1):
        block.append(name)
        p2 = pos2[name]
        if p2 > max_pos2:
            max_pos2 = p2

        if max_pos2 == i:
            # Internal order is unspecified by the rule -> sort alphabetically.
            block.sort()
            answer.extend(block)
            block.clear()

    # Output the final ranking (bytes -> bytes, fast)
    sys.stdout.buffer.write(b"\n".join(answer))

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