<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

175. Encoding
time limit per test: 0.25 sec.
memory limit per test: 4096 KB
input: standard
output: standard



Let phi(W) is the result of encoding for algorithm:
1. If the length of W is 1 then phi(W) is W;
2. Let coded word is W = w1w2...wN and K = N / 2 (rounded down);
3. phi(W) = phi(wNwN-1...wK+1) + phi(wKwK-1...w1).
For example, phi('Ok') = 'kO', phi('abcd') = 'cdab'.
Your task is to find position of letter wq in encoded word phi(W).

Input
Given integers N, q (1 <= N <= 10^9; 1<= q <= N), where N is the length of word W.

Output
Write position of letter wq in encoded word phi(W).

Sample test(s)

Input
9 4

Output
8
Author:	Michael R. Mirzayanov
Resource:	Saratov ST team Spring Contest #1
Date:	18.05.2003

<|response|>
1. Abridged problem statement  
Given an (unknown) string W of length N, define its encoded form φ(W) by:  
- If N=1, φ(W)=W.  
- Otherwise let K=⌊N/2⌋. Split W into A=w₁…w_K and B=w_{K+1}…w_N, reverse both halves, encode each recursively, and concatenate:  
  φ(W)=φ(reverse(B)) ‖ φ(reverse(A)).  
Given N and an index q (1≤q≤N), determine at which position in φ(W) the original character w_q ends up.

2. Key observations  
- We never need the actual letters; we only track how an index p moves under φ.  
- Each step splits the range [1..N] into first half A of size K=⌊N/2⌋ and second half B of size L=N−K.  
- Characters from B go first (after reversing B), then characters from A (after reversing A).  
- Reversal maps a local index p′ to new index (length − p′ + 1).  
- After reversal, we recurse on a smaller length (L or K).  
- This leads to a O(log N) recursion which is efficient for N up to 10⁹.

3. Full solution approach  
Define a function rec(n, p) that returns the final position of the original character at index p in a string of length n after encoding by φ.  

Base case:  
- If n=1, then rec(1, 1)=1.  

Recursive step (n>1):  
- Let K=⌊n/2⌋, L=n−K.  
- If p≤K (p lies in the first half A):  
  • After reversing A, its local index becomes r=K−p+1.  
  • That block encodes to a block of size K but is placed after the block of size L, so the final position is L + rec(K, r).  
- Else (p>K, p lies in the second half B):  
  • Local index in B is p−K; after reversal its index becomes r=(p−K) reversed within length L ⇒ r=L−(p−K)+1 = n−p+1.  
  • That block encodes to a block of size L placed first, so the final position is rec(L, r).

Answer = rec(N, q).

4. C++ implementation with detailed comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// rec(n, pos): returns where the original character at index pos
// in a length-n string ends up after encoding.
long long rec(long long n, long long pos) {
    // Base case: single character stays at position 1
    if (n == 1) {
        return 1;
    }
    // Split into first half size K and second half size L
    long long K = n / 2;
    long long L = n - K;

    if (pos <= K) {
        // Case 1: pos in first half A
        // After reversing A, local index becomes K - pos + 1
        long long r = K - pos + 1;
        // That part is encoded in length K and placed after the L-length block
        return L + rec(K, r);
    } else {
        // Case 2: pos in second half B
        // After reversing B, local index becomes n - pos + 1
        long long r = n - pos + 1;
        // That part is encoded in length L and placed first
        return rec(L, r);
    }
}

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

    long long N, q;
    cin >> N >> q;
    cout << rec(N, q) << "\n";
    return 0;
}
```

5. Python implementation with detailed comments  
```python
import sys
sys.setrecursionlimit(10**7)

def rec(n, pos):
    """
    Return the final position of the original character at index pos
    in a length-n string after applying the encoding φ.
    """
    # Base case: single character
    if n == 1:
        return 1
    # Split sizes
    K = n // 2       # size of first half A
    L = n - K        # size of second half B

    if pos <= K:
        # pos is in A
        # after reversal of A, new index is K - pos + 1
        r = K - pos + 1
        # this block of size K goes after the block of size L
        return L + rec(K, r)
    else:
        # pos is in B
        # after reversal of B, new index is n - pos + 1
        r = n - pos + 1
        # this block of size L goes first
        return rec(L, r)

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

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