1. Abridged Problem Statement
- There are two teams (Berland A and Byteland B) with n players each, all 2n skill levels are distinct, and the stronger player always wins.
- A sortition is a perfect matching between A and B (each A plays exactly one B).
- Count the number of sortitions where Berland wins exactly k more games than Byteland, modulo 1,000,000,009.

2. Detailed Editorial
- Let wins(A) and wins(B) be the wins of A and B in the final matching. We have:
  - wins(A) + wins(B) = n
  - wins(A) - wins(B) = k
  - Hence wins(B) = (n - k) / 2, wins(A) = (n + k) / 2. If n - k is odd, the answer is 0.

- Key observation via sorting:
  - Put all 2n players into one array as (skill, team), team = 0 for A and 1 for B, and sort by skill ascending.
  - Any match pairs a weaker earlier player with a stronger later player.
  - When we scan left-to-right, the only way to form a pair at position i is to connect the current player (as the stronger endpoint) with some unmatched earlier player of the opposite team.

- DP state while scanning the sorted array:
  - Let cnt_a, cnt_b be the counts of A and B seen so far in the prefix (before the current element).
  - Let dp[wa][wb] be the number of ways for the processed prefix to have already formed wa A-wins and wb B-wins. Each formed pair consumes exactly one A and one B from the prefix. Thus the number of unmatched earlier A's is cnt_a - (wa + wb), and unmatched earlier B's is cnt_b - (wa + wb).

- Transition on current element of team T:
  - Option 1: Leave current player unmatched (to be the weaker in a future pair).
    - dp'[wa][wb] += dp[wa][wb]
  - Option 2: Match current player now (as the stronger) with an earlier unmatched opponent:
    - If T = A, we can match with any unmatched B: candidates = cnt_b - (wa + wb). For each, we form an A-win:
      - dp'[wa + 1][wb] += dp[wa][wb] * candidates
    - If T = B, candidates = cnt_a - (wa + wb), and we form a B-win:
      - dp'[wa][wb + 1] += dp[wa][wb] * candidates
  - Move to the next element; finally increment cnt_a or cnt_b according to the current element's team.

- Why candidates = cnt_opposite - (wa + wb):
  - After forming wa + wb pairs so far, exactly wa + wb A's and wa + wb B's have already been used in those pairs (each pair consumes one A and one B).
  - So among all earlier opponents of the opposite team, only cnt_opposite - (wa + wb) remain available.

- Answer:
  - After processing all 2n players, the count is dp[wins(A)][wins(B)] modulo 1e9+9.

- Complexity:
  - There are 2n steps. The DP table has size (wins(A)+1) × (wins(B)+1) = O(n^2).
  - Total time O(n · wins(A) · wins(B)) = O(n^3) in worst case; memory O(wins(A) · wins(B)).
  - With n ≤ 500, this is fine in C++.

3. C++ Solution

```cpp
#include <bits/stdc++.h>
#include <vector>

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;
};

const int mod = (int)1e9 + 9;

void mod_add(int& a, int b) {
    a += b;
    if(a >= mod) {
        a -= mod;
    }
}

int n, k;
vector<int> a, b;

void read() {
    cin >> n >> k;
    a.resize(n);
    b.resize(n);
    cin >> a >> b;
}

void solve() {
    // We want the number of wins by team A to be k more
    // than the number of wins of team B. We know that
    // wins(A) + wins(B) = n, so 2 * wins(B) + k = n, or
    // wins(B) = (n - k) / 2. The constraints in this problem
    // should immediately lead us to think about DP solutions.
    // Before we start with the actual idea, lets sort the individual
    // strengths (the a and b arrays), as we are anyways considering
    // permutations. In particular, we will create a joint sequence
    // of length 2 * n, that will contain all people and sort it. Then
    // we are essentially interested in ways of matching up positions
    // of this joint sequence such that we have exactly wins(B). To do
    // this we will consider the following DP state:
    //
    //    dp[prefix][wins(A) so far][wins(B) so far]
    //      - We are looking at prefix of the corresponding length.
    //      - Team A has claimed wins(A) wins from complete matches.
    //      - Team B has claimed wins(B) wins from complete matches.
    //
    // In particular, when we are at position i, we can either match
    // an existing pair, or start a new one. The state has enough
    // information for us to determine the number of unmatched
    // As and Bs - in particular we have wins(A) + wins(B) matched
    // elements of each A and B, and we know how many As and Bs there
    // are in the current prefix. Note that we multiply by the number of
    // valid candidates only when we "finish" the pair to avoid over counting.
    //
    // This has O(N^3) time complexity, but we can notice that we can only keep
    // two rows in the first dimension.

    int win_b = (n - k);
    if(win_b % 2 == 1) {
        cout << 0 << endl;
        return;
    }

    win_b /= 2;
    int win_a = win_b + k;

    vector<pair<int, int>> elements;
    for(int i = 0; i < n; i++) {
        elements.push_back({a[i], 0});
        elements.push_back({b[i], 1});
    }

    sort(elements.begin(), elements.end());

    int cnt_a = 0, cnt_b = 0;
    vector<vector<int>> dp(win_a + 1, vector<int>(win_b + 1, 0));
    dp[0][0] = 1;

    for(auto [val, type]: elements) {
        vector<vector<int>> new_dp(win_a + 1, vector<int>(win_b + 1, 0));
        for(int curr_win_a = 0; curr_win_a <= win_a; curr_win_a++) {
            for(int curr_win_b = 0; curr_win_b <= win_b; curr_win_b++) {
                if(!dp[curr_win_a][curr_win_b]) {
                    continue;
                }

                mod_add(
                    new_dp[curr_win_a][curr_win_b], dp[curr_win_a][curr_win_b]
                );

                int new_win_a = curr_win_a, new_win_b = curr_win_b;
                int candidates = -(curr_win_a + curr_win_b);
                if(type == 0) {
                    candidates += cnt_b;
                    new_win_a++;
                } else {
                    candidates += cnt_a;
                    new_win_b++;
                }

                if(candidates < 0) {
                    continue;
                }

                if(new_win_a <= win_a && new_win_b <= win_b) {
                    mod_add(
                        new_dp[new_win_a][new_win_b],
                        dp[curr_win_a][curr_win_b] * 1ll * candidates % mod
                    );
                }
            }
        }

        dp = std::move(new_dp);
        if(type == 0) {
            cnt_a++;
        } else {
            cnt_b++;
        }
    }

    cout << dp[win_a][win_b] << 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

```python
import sys

MOD = 1_000_000_009

def solve():
    data = sys.stdin.read().strip().split()
    it = iter(data)
    n = int(next(it))
    k = int(next(it))
    a = [int(next(it)) for _ in range(n)]
    b = [int(next(it)) for _ in range(n)]

    # wins(B) = (n - k) / 2 must be integer, else 0
    if (n - k) % 2 != 0:
        print(0)
        return

    win_b = (n - k) // 2
    win_a = win_b + k  # (n + k) // 2

    # Build combined list: (strength, team), team: 0 = A (Berland), 1 = B (Byteland)
    elements = [(x, 0) for x in a] + [(x, 1) for x in b]
    elements.sort()  # increasing by strength

    # dp[wa][wb] = ways after processing the current prefix
    # to have formed wa A-wins and wb B-wins.
    dp = [[0] * (win_b + 1) for _ in range(win_a + 1)]
    dp[0][0] = 1

    cnt_a = 0  # how many A's seen so far (in the prefix before current element)
    cnt_b = 0  # how many B's seen so far

    for value, team in elements:
        # new dp after incorporating this current element
        new_dp = [[0] * (win_b + 1) for _ in range(win_a + 1)]

        # Traverse all states of formed wins so far
        for wa in range(win_a + 1):
            row = dp[wa]
            new_row_same = new_dp[wa]  # small speed-up by caching
            for wb in range(win_b + 1):
                ways = row[wb]
                if ways == 0:
                    continue

                # Option 1: leave current element unmatched
                new_row_same[wb] = (new_row_same[wb] + ways) % MOD

                # Option 2: match current (as stronger) with an earlier unmatched opponent
                # Number of earlier unmatched opponents of the opposite team:
                # equals (count_opposite_in_prefix) - (#pairs_already_closed),
                # where #pairs_already_closed = wa + wb.
                if team == 0:
                    # current is A, can close with B
                    candidates = cnt_b - (wa + wb)
                    if candidates > 0 and wa + 1 <= win_a:
                        new_dp[wa + 1][wb] = (new_dp[wa + 1][wb] + ways * candidates) % MOD
                else:
                    # current is B, can close with A
                    candidates = cnt_a - (wa + wb)
                    if candidates > 0 and wb + 1 <= win_b:
                        new_dp[wa][wb + 1] = (new_dp[wa][wb + 1] + ways * candidates) % MOD

        dp = new_dp

        # Update counts of seen players for the next iteration
        if team == 0:
            cnt_a += 1
        else:
            cnt_b += 1

    print(dp[win_a][win_b] % MOD)

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

Note: The algorithm is O(n * win_a * win_b) ≈ O(n^3) in the worst case; this is easily fast in C++, but Python may be slow at the upper limits. The implementation is faithful to the intended solution.

5. Compressed Editorial
- Sort all players by strength; a match always connects an earlier weaker to a later stronger.
- Scan left-to-right. Maintain dp[wa][wb] = ways to have already formed wa A-wins and wb B-wins in the prefix, and counts cnt_a, cnt_b of A/B seen so far.
- For the current player:
  - Leave it unmatched: dp'[wa][wb] += dp[wa][wb].
  - Or match it now (as stronger) with an earlier unmatched opponent of the other team:
    - Available opponents = (cnt_opposite − (wa + wb)), because wa + wb pairs have already consumed that many players from each team.
    - If current is A: dp'[wa+1][wb] += dp[wa][wb] * candidates. If B: dp'[wa][wb+1] += dp[wa][wb] * candidates.
- wins(B) = (n − k)/2 must be integral; answer is dp[(n + k)/2][(n − k)/2] modulo 1e9+9.
- Complexity: O(n^3) time, O(n^2) memory; passes for n ≤ 500 in C++.
