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. 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;

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

int rec(int len, int pos) {
    if(len == 1) {
        return 1;
    }

    int k = len / 2;
    if(pos <= k) {
        return len - k + rec(k, k - pos + 1);
    }

    return rec(len - k, len - pos + 1);
}

void solve() {
    // phi splits W into the second half (reversed) followed by the first
    // half (reversed), recursively. Mirror the same recursion to locate where
    // letter w_q lands: with k = len / 2, position pos in the first half maps
    // into the encoded second block (offset by len - k) at mirrored position
    // k - pos + 1, while a position in the second half maps into the first
    // block at mirrored position len - pos + 1. Recurse until length 1.

    cout << rec(n, q) << '\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(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).