1. Abridged Problem Statement
Given N numbers (1…N) arranged in a circle. Start at 1, moving clockwise. Repeat until all numbers are removed:
  • Count Q numbers in your current direction (including the starting position as 1), erase that Q-th number.
  • Move one step clockwise from the erased position.
  • If you land on an odd number, set direction = clockwise; if even, set direction = counter-clockwise.
Output the last erased number.

2. Detailed Editorial
We must simulate a Josephus-style removal with a twist: the counting direction can flip after each removal, based on the parity of the next number you land on. A direct simulation maintaining the circle as a linked list of size up to N=2 000 000 is feasible in O(N·Q) time since Q≤10 and N·Q≈2×10^7.

Data Structures:
  • Two arrays (or vectors) nxt[1..N], prv[1..N] implementing a doubly linked circular list.
     – nxt[i] = the label of the node clockwise from i
     – prv[i] = the label of the node counter-clockwise from i

Initialization:
  For i from 1 to N:
    nxt[i] = i+1 (or 1 if i==N)
    prv[i] = i−1 (or N if i==1)

Simulation Variables:
  • current = 1 (your current position)
  • clockwise = true (current direction)
  • remaining = N (nodes left)
  • last_erased = −1

Loop while remaining>0:
  1. Let pos = current
  2. Move (Q−1) steps in the current direction:
       repeat Q−1 times:
         if clockwise: pos = nxt[pos]
         else:        pos = prv[pos]
  3. Erase pos:
       – link prv[pos]→nxt[pos] and nxt[pos]→prv[pos]
       – remaining−−
       – last_erased = pos
  4. If remaining==0 break.
  5. Move current = nxt[pos]  (always one step clockwise after erasure).
  6. Update direction: clockwise = (current is odd)

After the loop, last_erased holds the answer.

Time Complexity: O(N·Q). Memory: O(N).

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, q;
vector<int> nxt, prv;

void read() { cin >> n >> q; }

void solve() {
    // Direct simulation on a doubly linked list over the circle 1..N, where
    // nxt[i] / prv[i] are the clockwise / counter-clockwise neighbours.
    // Starting at 1 moving clockwise, each step walks Q positions in the
    // current direction to the number to erase, unlinks it, then advances one
    // step clockwise. The parity of the number landed on after that step sets
    // the direction for the next step (odd -> clockwise, even ->
    // counter-clockwise). We keep going until the list is empty and report the
    // last erased number.

    nxt.resize(n + 1);
    prv.resize(n + 1);
    for(int i = 1; i <= n; i++) {
        nxt[i] = (i == n) ? 1 : i + 1;
        prv[i] = (i == 1) ? n : i - 1;
    }

    int* nx = nxt.data();
    int* pv = prv.data();
    int current = 1;
    bool clockwise = true;
    int last_erased = -1;

    for(int remaining = n; remaining > 0; remaining--) {
        int pos = current;
        if(clockwise) {
            for(int i = 1; i < q; i++) {
                pos = nx[pos];
            }
        } else {
            for(int i = 1; i < q; i++) {
                pos = pv[pos];
            }
        }

        last_erased = pos;

        int next_pos = nx[pos];
        nx[pv[pos]] = next_pos;
        pv[next_pos] = pv[pos];

        current = next_pos;
        clockwise = (current & 1);
    }

    cout << last_erased << '\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
import sys
sys.setrecursionlimit(10**7)
def main():
    data = sys.stdin.read().split()
    N, Q = map(int, data)
    # Use 1-based indexing; index 0 unused
    nxt = [i+1 for i in range(N+1)]
    prv = [i-1 for i in range(N+1)]
    nxt[N] = 1
    prv[1] = N

    current = 1          # start at node 1
    clockwise = True     # initial direction
    remaining = N
    last_erased = -1

    # Loop until all nodes removed
    while remaining > 0:
        # Step 1: find Q-th node in current direction
        pos = current
        for _ in range(Q-1):
            pos = nxt[pos] if clockwise else prv[pos]

        # Step 2: erase pos from the circle
        last_erased = pos
        left = prv[pos]
        right = nxt[pos]
        nxt[left] = right
        prv[right] = left
        remaining -= 1

        if remaining == 0:
            break

        # Step 3: move one step clockwise
        current = right

        # Step 4: update direction by parity of current
        if current & 1:
            clockwise = True
        else:
            clockwise = False

    # Output the last erased number
    print(last_erased)

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

5. Compressed Editorial
Use a circular doubly linked list (arrays nxt, prv). Track current position and a boolean for direction. Each iteration: step Q−1 times in direction, remove that node by relinking neighbors, move one clockwise step, then set direction = clockwise if that node is odd else counter-clockwise. Repeat until empty; the last removed is the answer.
