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

// Overload output for pair<T1,T2>
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload input for pair<T1,T2>
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload input for vector<T>
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Overload output for vector<T>
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;

// Read input N and Q
void read() {
    cin >> N >> Q;
}

// Solve one test case
void solve() {
    // Allocate linked-list arrays of size N+1 (1-based)
    nxt.resize(N + 1);
    prv.resize(N + 1);

    // Initialize the circular doubly linked list
    for(int i = 1; i <= N; i++) {
        nxt[i] = (i == N) ? 1 : (i + 1);  // next clockwise
        prv[i] = (i == 1) ? N : (i - 1);  // previous (counter-clockwise)
    }

    int current = 1;       // start at node 1
    bool clockwise = true; // initial direction is clockwise
    int last_erased = -1;
    int remaining = N;     // how many nodes still in the circle

    // Main simulation loop
    while(remaining > 0) {
        // 1) Find the Q-th node in the current direction
        int pos = current;
        for(int i = 1; i < Q; i++) {
            if(clockwise) {
                pos = nxt[pos];
            } else {
                pos = prv[pos];
            }
        }

        // 2) Erase pos
        last_erased = pos;
        int next_pos = nxt[pos];
        int prev_pos = prv[pos];
        // unlink pos
        nxt[prev_pos] = next_pos;
        prv[next_pos] = prev_pos;
        remaining--;

        // If that was the last, we're done
        if(remaining == 0) {
            break;
        }

        // 3) Move one step clockwise from the erased node
        current = next_pos;

        // 4) Update direction by parity of the node we landed on
        if(current % 2 == 1) {
            clockwise = true;
        } else {
            clockwise = false;
        }
    }

    // Output the last erased number
    cout << last_erased << "\n";
}

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

    int T = 1; 
    // If multiple tests, read T
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        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.