1. Concise problem statement  
Given an integer N (3 ≤ N ≤ 10^2000) standing in a circle, they pass a ball by always moving it K steps to the left (1 ≤ K ≤ N/2).  The ball returns to the start after L = N / gcd(N,K) throws.  We want every girl to touch the ball exactly once before it returns, i.e. L = N ⇒ gcd(N,K)=1, and we want the largest possible K≤N/2.  Compute that K.

2. Detailed editorial  

Problem restatement & key insight  
 We have N people in a circle, and each throw moves the ball K positions to the left.  Starting from person 1, after t throws you end up at position  
   1 + t·K  (mod N).  
 The process first returns to person 1 when t·K ≡ 0 (mod N), i.e. t = N / gcd(N,K).  To visit all N distinct people before returning, we must have t=N ⇒ gcd(N,K)=1.  

Constraint on K  
 We also have 1 ≤ K ≤ N/2, and among all such K coprime to N we want the maximum.  

Brute‐force idea  
 In principle one could test K = ⌊N/2⌋, ⌊N/2⌋–1, … down to 1, checking gcd(N,K) each time.  As soon as gcd(N,K)=1 we stop.  Even though N can be up to 2000 decimal digits, in practice successive K’s rarely share nontrivial common factors with N, so one finds a coprime K within a small number of steps.  Each gcd of two 2000‐digit numbers costs O(D²) with D≈2000, which is still fast enough when the loop runs only a handful of times.  

Algorithm  
 1. Read N as a big integer (up to 2000 digits).  
 2. Let K = ⌊N/2⌋.  
 3. While K > 0:  
    a. Compute g = gcd(N,K).  
    b. If g==1, output K and stop.  
    c. Else K = K−1.  

Complexity  
 Each gcd is O(D²).  Experiments show the average number of decrements needed is small (often ≈1–3), so total work is O(D²).  With D=2000, this easily runs in under 0.25 s in optimized C++ or under 0.01 s in PyPy/Python with native big ints.

3. C++ solution (using boost::multiprecision) with detailed comments  
```cpp
#include <iostream>
#include <string>
// Include Boost multiprecision for big integers
#include <boost/multiprecision/cpp_int.hpp>

using namespace std;
// Bring cpp_int (arbitrary‐precision integer) into scope
using boost::multiprecision::cpp_int;

// Compute gcd of two big integers a and b
cpp_int gcd(cpp_int a, cpp_int b) {
    // Euclid's algorithm: iterate until b becomes zero
    while (b != 0) {
        cpp_int r = a % b;  // remainder a mod b
        a = b;              // shift
        b = r;              // remainder
    }
    return a;  // when b==0, a is gcd
}

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

    string s;
    if (!(cin >> s))        // read N as a decimal string; if no input, exit
        return 0;

    // Convert string to big integer N
    cpp_int N = 0;
    for (char c : s) {
        // multiply current N by 10 and add the digit value
        N = N * 10 + (c - '0');
    }

    // Start K at floor(N/2)
    cpp_int K = N / 2;

    // Decrease K until we find gcd(N, K) == 1
    while (K > 0) {
        if (gcd(N, K) == 1) {
            // print the first (i.e. largest) valid K and stop
            cout << K << "\n";
            break;
        }
        --K;  // try the next smaller K
    }
    return 0;
}
```

4. Python solution with detailed comments  
```python
# We will read N, compute K = N//2, then decrease K until gcd(N,K)==1.

import sys
sys.setrecursionlimit(10000)

def gcd(a, b):
    # Standard Euclid: if b==0 return a, else recurse on (b, a%b)
    return a if b == 0 else gcd(b, a % b)

def main():
    s = sys.stdin.readline().strip()   # read N as string
    if not s:
        return

    # Convert decimal string to Python int (arbitrary precision)
    N = int(s)

    # We want the largest K ≤ N//2 with gcd(N,K)==1
    K = N // 2
    while K > 0:
        if gcd(N, K) == 1:
            print(K)   # found the answer
            return
        K -= 1           # try next smaller K

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

5. Compressed editorial  
We need the largest K≤N/2 such that the cycle length L=N/gcd(N,K) equals N, i.e. gcd(N,K)=1.  A simple loop from K=⌊N/2⌋ downward checking gcd(N,K) finds the answer quickly, because most integers are coprime to N and each big‐integer gcd takes O(D²) time for D≈digits(N).