1) Abridged problem statement
- We write n letters arranged in a circle, each letter being either X or E.
- Two strings are considered the same if one can be rotated to obtain the other (rotations only; no reflections).
- Given 1 ≤ n ≤ 200000, compute the number of distinct circular strings (binary necklaces) of length n.

2) Detailed editorial
- Model: We are counting binary necklaces of length n under the action of the cyclic group C_n (rotations).
- Tool: Burnside’s Lemma (Cauchy-Frobenius) says the number of distinct objects (orbits) is the average number of configurations fixed by each group action (rotation).

- Fixed configurations under a rotation:
  - Consider rotation by k positions (k = 0..n-1). This rotation partitions the n positions into gcd(n, k) cycles. Each cycle must be constant in any fixed string, and each cycle can be chosen as X or E independently. Therefore the number of binary strings fixed by rotation k is 2^{gcd(n, k)}.

- Burnside summation:
  - Answer = (1/n) * sum_{k=0}^{n-1} 2^{gcd(n, k)}.

- Grouping by gcd via Euler’s totient:
  - The number of k with gcd(n, k) = d (for d | n) equals φ(n/d). Hence we can rewrite:
    Answer = (1/n) * sum_{d | n} φ(n/d) * 2^d
    which is equivalent to:
    Answer = (1/n) * sum_{d | n} φ(d) * 2^{n/d}

- Complexity and implementation:
  - n ≤ 200000, so both of these are efficient:
    - Direct sum over k: O(n log n) (due to gcd), perfectly fine in Python.
    - Divisor/totient form: O(τ(n) + factoring), with τ(n) the number of divisors (≤ ~1500 here).
  - Big integers are required since the answer is roughly 2^n / n; Python handles this natively; in C++ use boost::multiprecision::cpp_int.
  - Implementation plan (divisor/totient form):
    1) Factor n to get its distinct prime divisors.
    2) Enumerate all divisors d | n.
    3) For each m = d (or m = n/d, depending on the chosen formula), compute φ(m) using the primes of n:
       φ(m) = m * Π_{p | m} (1 - 1/p).
    4) Accumulate φ(d) * 2^{n/d} (or φ(n/d) * 2^d).
    5) Divide the sum by n and print.

- Correctness check (samples):
  - n = 3: (1/3)(2^3 + 2^1 + 2^1) = 4.
  - n = 4: (1/4)(2^4 + 2^1 + 2^2 + 2^1) = 6.

3) C++ solution (with detailed comments)
Note: No C++ file was provided in the bundle; below is a correct and fully commented C++ implementation using Boost big integers.

```cpp
// We use a totient-grouped Burnside formula:
//   answer = (1/n) * sum_{d | n} phi(d) * 2^(n/d)
//
// Big integers are necessary because the result can be very large.
// We use boost::multiprecision::cpp_int for arbitrary precision.

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>

using namespace std;
// Big integer type.
using boost::multiprecision::cpp_int;

// Return the distinct prime factors of x.
// We only need distinct primes, not their exponents, because Euler's phi
// only depends on which primes divide the number.
static vector<int> distinct_prime_factors(int x) {
    vector<int> primes;
    // Work on a copy that we can divide.
    int n = x;

    // Check factor 2 separately.
    if (n % 2 == 0) {
        primes.push_back(2);
        while (n % 2 == 0) n /= 2;
    }

    // Check odd factors up to sqrt(n).
    for (int p = 3; 1LL * p * p <= n; p += 2) {
        if (n % p == 0) {
            primes.push_back(p);
            while (n % p == 0) n /= p;
        }
    }

    // If there's a prime factor > sqrt(original x), it's n itself > 1.
    if (n > 1) primes.push_back(n);

    return primes;
}

// Enumerate all positive divisors of n.
static vector<int> divisors_of(int n) {
    vector<int> divs;
    for (int i = 1; 1LL * i * i <= n; ++i) {
        if (n % i == 0) {
            divs.push_back(i);
            if (i != n / i) divs.push_back(n / i);
        }
    }
    return divs;
}

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

    int n;
    // Read n; if no input, exit.
    if (!(cin >> n)) return 0;

    // Get distinct prime divisors of n to help compute φ for any divisor of n.
    vector<int> primes = distinct_prime_factors(n);

    // Enumerate all divisors d of n.
    vector<int> divs = divisors_of(n);

    cpp_int sum = 0;  // Will hold the sum of phi(d) * 2^(n/d) over all d | n.

    for (int d : divs) {
        // We need phi(d). Since d | n, the primes dividing d are a subset of primes.
        long long m = d;       // We'll compute phi(d) in a 64-bit integer (fits since d <= 200000).
        long long phi = m;     // Start from phi(d) = d.

        // Reduce phi by each prime dividing d: phi = phi / p * (p - 1).
        for (int p : primes) {
            if (m % p == 0) {
                phi = phi / p * (p - 1);
                // Remove all factors of p from m (not necessary for correctness,
                // but cheap and shows we've accounted for this prime).
                while (m % p == 0) m /= p;
            }
        }
        // If m > 1 here, it would be a prime factor not in 'primes' (shouldn't happen),
        // but since primes are the distinct primes of n and d | n, m should become 1.

        // Compute term = phi(d) * 2^(n/d).
        // 2^(n/d) is a large number; use big integer shifts for fast power-of-two.
        int exponent = n / d;
        cpp_int pow2 = cpp_int(1) << exponent;  // 1 shifted left by exponent bits = 2^exponent.
        cpp_int term = cpp_int(phi) * pow2;

        sum += term;  // Accumulate into big integer.
    }

    // Finally divide by n according to Burnside.
    cpp_int answer = sum / n;

    // Output the big integer.
    cout << answer << '\n';
    return 0;
}
```

4) Python solution (with detailed comments)
Below is a slightly optimized version using the divisor/totient grouping; Python’s big ints handle the size naturally.

```python
import sys
from math import isqrt

# On Python 3.11+, printing extremely large integers may require raising this limit.
if hasattr(sys, "set_int_max_str_digits"):
    try:
        sys.set_int_max_str_digits(10**7)
    except Exception:
        pass

def distinct_prime_factors(x: int) -> list[int]:
    """
    Return the list of distinct prime divisors of x.
    We only need distinct primes (no multiplicities) to compute Euler's totient.
    """
    primes = []
    n = x

    # Factor out 2.
    if n % 2 == 0:
        primes.append(2)
        while n % 2 == 0:
            n //= 2

    # Factor out odd primes up to sqrt(n).
    p = 3
    while p * p <= n:
        if n % p == 0:
            primes.append(p)
            while n % p == 0:
                n //= p
        p += 2

    # If there's a leftover prime factor > 1, it's prime.
    if n > 1:
        primes.append(n)

    return primes

def divisors_of(n: int) -> list[int]:
    """
    Return the list of all positive divisors of n.
    """
    divs = []
    r = isqrt(n)
    for i in range(1, r + 1):
        if n % i == 0:
            divs.append(i)
            j = n // i
            if j != i:
                divs.append(j)
    return divs

def euler_phi_of_divisor_using_n_primes(d: int, primes_of_n: list[int]) -> int:
    """
    Compute φ(d) for d | n, given the distinct prime divisors of n.
    Since primes_of_n contains all primes that could divide d, we can compute φ(d)
    by checking only these primes (others can't divide d).
    """
    phi = d
    m = d
    for p in primes_of_n:
        if m % p == 0:
            phi = phi // p * (p - 1)
            while m % p == 0:
                m //= p
    # m should be 1 here if d | n; otherwise there would be an extra prime not in primes_of_n.
    return phi

def solve():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    n = int(data[0])

    # Precompute distinct prime divisors of n for φ computations.
    primes = distinct_prime_factors(n)

    # Enumerate divisors d of n.
    divs = divisors_of(n)

    # Sum φ(d) * 2^(n/d) using big integers (Python ints are arbitrary precision).
    total = 0
    for d in divs:
        phi_d = euler_phi_of_divisor_using_n_primes(d, primes)
        exponent = n // d
        # 2^(n/d) can be computed efficiently as 1 << exponent (bit-shift).
        total += phi_d * (1 << exponent)

    # Final answer by Burnside: divide by n.
    ans = total // n
    print(ans)

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

5) Compressed editorial
- We count binary necklaces (rotations only) of length n.
- Burnside’s Lemma: the number of distinct necklaces equals the average number of strings fixed by each rotation.
- A rotation by k fixes exactly 2^{gcd(n, k)} strings because the positions split into gcd(n, k) cycles, each freely chosen as X or E.
- Answer = (1/n) * sum_{k=0}^{n-1} 2^{gcd(n, k)}. Group by gcd values to get:
  Answer = (1/n) * sum_{d | n} φ(d) * 2^{n/d}.
- Implement by enumerating divisors of n, computing φ(d) via the primes of n, summing φ(d) * 2^{n/d}, and dividing by n. Use big integers.