1. Abridged Problem Statement  
Given a word W of length N, define its encoding φ(W) recursively as follows:  
- If N=1, φ(W)=W.  
- Otherwise, let K=⌊N/2⌋. Split W into first half A=w₁…w_K and second half B=w_{K+1}…w_N. Reverse B and A, encode each recursively, and concatenate:  
  φ(W)=φ(reverse(B)) ‖ φ(reverse(A)).  
Given N and an index q (1≤q≤N), determine the position of the original character w_q in the encoded word φ(W).

2. Detailed Editorial  
We do not need the actual characters—only track how positions move under φ. Define a function rec(n, p) = new position of original index p in φ applied to a length-n string.

Base case:  
- n=1 ⇒ rec(1,1)=1.

Recursive step (n>1):  
- Let K=⌊n/2⌋, L=n−K. In φ(W), the first L positions come from encoding of reversed second half, and the next K positions come from encoding of reversed first half.

Case 1: p > K (p in second half B of length L)  
  - Within B before reversal, its local index is p' = p−K (1≤p'≤L). After reversing B, that character moves to reversed index r = L−p'+1 = n−p+1.  
  - Then it goes through φ on length L, so its final position is rec(L, r).

Case 2: p ≤ K (p in first half A of length K)  
  - Local index p in A; after reversing A its index is r = K−p+1.  
  - Then it goes through φ on length K, yielding rec(K, r), but that block is placed after the block of size L, so we add L: final = L + rec(K, r) = (n−K) + rec(K, K−p+1).

Since each recursive call halves n, the time complexity is O(log N). N can be up to 10⁹, so this is efficient.

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

// Global variables to hold inputs
int n, q;

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

// rec(n, pos) returns the position of the original character at index pos
// after encoding a length-n string.
int rec(int n, int pos) {
    // Base case: a single character maps to position 1
    if (n == 1) {
        return 1;
    }
    // K = size of the first half, L = size of the second half
    int k = n / 2;
    int l = n - k;
    
    // If pos is in the first half (A)
    if (pos <= k) {
        // After reversing A, local index becomes k - pos + 1
        int newPosInA = k - pos + 1;
        // We then encode that half (size k), and shift by l in the final output
        return l + rec(k, newPosInA);
    } else {
        // pos is in the second half (B)
        // After reversing B, local index becomes (n - pos + 1)
        int newPosInB = n - pos + 1;
        // We then encode that half (size l) and place it at the front: no shift
        return rec(l, newPosInB);
    }
}

void solve() {
    // Compute and output the final mapped position
    cout << rec(n, q) << '\n';
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
sys.setrecursionlimit(1000000)

def rec(n, pos):
    """
    Returns the position of the original character at index 'pos'
    after applying the recursive encoding to a length-n string.
    """
    # Base case: single character remains at position 1
    if n == 1:
        return 1
    # Split point
    k = n // 2        # size of first half A
    l = n - k         # size of second half B

    if pos <= k:
        # Character is in the first half A
        # After reversing A, its index becomes (k - pos + 1)
        new_pos = k - pos + 1
        # It is encoded in the second part of the result,
        # so we add l to the recursive result
        return l + rec(k, new_pos)
    else:
        # Character is in the second half B
        # After reversing B, its index becomes (n - pos + 1)
        new_pos = n - pos + 1
        # It is encoded in the first part of the result
        return rec(l, new_pos)

def main():
    data = sys.stdin.read().split()
    N, q = map(int, data)
    print(rec(N, q))

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

5. Compressed Editorial  
Use a recursive function rec(n,p). If n=1 return 1. Split n into k=⌊n/2⌋, ℓ=n−k.  
- If p≤k: it lies in the first half; after reverse its index is k−p+1; final position = ℓ + rec(k, k−p+1).  
- Else: it lies in the second half; after reverse its index is n−p+1; final position = rec(ℓ, n−p+1).  
This runs in O(log N).