<|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.

246. Black & White
time limit per test: 0.25 sec.
memory limit per test: 65536 KB
input: standard
output: standard



Consider a necklace made of 2N-1 black and white beads, K of which are black. Necklace is called "beautiful" if it is possible to choose two black beads (not necessarily different) in such a way that one of two necklace parts strictly between them contains exactly N beads.
For example, if N=4 and K=3, necklace "WBWBWBW" is beautiful, and necklace "BBWWBWW" is not.
You need to find minimal K for which every necklace of 2N-1 beads is beatiful.

Input
The first line of input contains odd integer number 2N-1 (5<=2N-1<=2^31-1).

Output
Output minimal K for which every necklace of 2N-1 beads is beatiful.

Sample test(s)

Input
Test #1
5

Test #2
7

Output
Test #1
3

Test #2
4
Author:	Alexey Preobrajensky
Resource:	Petrozavodsk Summer Training Sessions 2004
Date:	August 25, 2004

<|response|>
1) Abridged problem statement
- You are given an odd integer L = 2N − 1 (5 ≤ L ≤ 2^31 − 1). Consider any necklace of L beads colored black or white.
- The necklace is “beautiful” if there exist two black beads (possibly the same bead) such that one of the two open arcs strictly between them contains exactly N beads.
- Find the minimal K such that every coloring with at least K black beads is necessarily beautiful.

2) Key observations
- Let two beads be separated by d steps along the circle (0 < d < L). The two open arcs strictly between them contain d − 1 and L − d − 1 beads.
- An arc has exactly N beads iff d − 1 = N or L − d − 1 = N, i.e., d = N + 1 or d = N − 2. These are the same undirected separation (one direction vs. the other), so it suffices to consider pairs at distance d = N + 1 (mod L).
- Build a graph G on the L positions; connect i to i + (N + 1) (mod L). A necklace is not beautiful iff the black positions form an independent set in G.
- The graph “jump by fixed step modulo L” is 2-regular and decomposes into g = gcd(L, N + 1) disjoint cycles, each of length L/g.
- A cycle of length len has maximum independent set size floor(len/2). Therefore the maximum number of black beads in a non-beautiful necklace is:
  α(G) = g * floor((L/g)/2)
- The minimal K that forces beauty for all colorings is α(G) + 1.

Useful simplification:
- gcd(2N − 1, N + 1) = gcd(N + 1, 3) ∈ {1, 3}.
- Hence:
  - If N ≡ 2 (mod 3): answer = N − 1
  - Otherwise: answer = N

3) Full solution approach
- Read L (the number of beads). Compute N = (L + 1) // 2.
- Either:
  - Use the simplified modular rule: output N − 1 if N % 3 == 2, else N; or
  - Use the general formula:
    - g = gcd(L, N + 1)
    - max_avoid = g * ((L // g) // 2)
    - answer = max_avoid + 1
- Both run in O(1) arithmetic plus O(log L) for gcd; memory O(1).

4) C++ implementation with detailed comments

```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;
};

int64_t n, two_n_minus_one;

void read() {
    cin >> two_n_minus_one;
    n = (two_n_minus_one + 1) / 2;
}

void solve() {
    // The necklace is beautiful iff there exist two black beads at distance
    // exactly n+1 (because then one of the open arcs has exactly n beads
    // strictly between them). The graph with edges i -- (i + (n+1)) mod (2n-1)
    // is a disjoint union of g = gcd(2n-1, n+1) cycles, each of length
    // (2n-1)/g. In a cycle of length len you can colour at most floor(len/2)
    // vertices black without creating a monochromatic edge. Hence the maximal
    // number of black beads without a beautiful configuration is g * floor(
    // ((2n-1)/g) / 2 ) and the minimal K that forces beauty for every possible
    // necklace is exactly one more.

    int64_t L = two_n_minus_one;
    int64_t dist = n + 1;
    int64_t g = gcd(L, dist);
    int64_t max_avoid = g * ((L / g) / 2);
    int64_t ans = max_avoid + 1;

    cout << ans << endl;
}

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;
}
```

5) Python implementation with detailed comments
import sys
import math

def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    L = int(data[0])              # L = 2N - 1

    N = (L + 1) // 2

    # Option A (closed form):
    # ans = N - (1 if N % 3 == 2 else 0)

    # Option B (gcd-based formula):
    d = N + 1
    g = math.gcd(L, d)
    max_avoid = g * ((L // g) // 2)
    ans = max_avoid + 1

    print(ans)

if __name__ == "__main__":
    main()