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. 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} 2^{gcd(n, k)}, where k runs over the n rotations.

- Two ways to evaluate the sum:
  - Direct bucketing by gcd value (used in the C++ solution): for each i in [1, n] compute g = gcd(n, i) and increment a counter cnt[g]. Since gcd(n, i) only takes divisor values of n, the sum becomes sum_{v} cnt[v] * 2^v. (Summing i over [1, n] is the same as over [0, n-1] because i = n and i = 0 both give g = n.) This is O(n log n) due to the gcd loop, perfectly fine for n ≤ 200000.
  - Grouping by gcd via Euler's totient (used in the Python solution): the number of k with gcd(n, k) = d (for d | n) equals φ(n/d). Hence
    Answer = (1/n) * sum_{d | n} φ(n/d) * 2^d = (1/n) * sum_{d | n} φ(d) * 2^{n/d}.
    This touches only τ(n) divisors and is even faster.

- Big integers and complexity:
  - The answer is roughly 2^n / n, so arbitrary-precision arithmetic is required. The C++ solution uses a vendored `bigint` struct (base-10^9 limbs with Karatsuba multiplication); Python handles big integers natively.
  - In the C++ solution, the numerator is accumulated as a bigint and divided exactly by n (an `int`) at the end.

- 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

```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;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

int n;

bigint fast_pow(int p, int k) {
    bigint r(1), b(p);
    while(k) {
        if(k & 1) {
            r *= b;
        }

        b *= b;
        k >>= 1;
    }

    return r;
}

void read() { cin >> n; }

void solve() {
    // Burnside's lemma over the cyclic group of circular shifts. A shift by i
    // fixes a binary string iff positions in each of its gcd(n, i) cycles agree,
    // so it fixes 2^gcd(n, i) strings. The number of distinct circular strings
    // is (sum over i of 2^gcd(n, i)) / n.
    //
    // gcd(n, i) only takes the divisor values of n, so we bucket how many i in
    // [1, n] yield each gcd value and weight 2^v by that count. The numerator is
    // computed with the vendored bigint and divided exactly by n at the end.

    vector<int> cnt(n + 1, 0);
    for(int i = 1; i <= n; i++) {
        int a = n, b = i;
        while(b) {
            a %= b;
            swap(a, b);
        }

        cnt[a]++;
    }

    bigint ans(0);
    for(int v = 1; v <= n; v++) {
        if(cnt[v] == 0) {
            continue;
        }

        ans += fast_pow(2, v) * cnt[v];
    }

    ans /= n;
    cout << ans << '\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
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} 2^{gcd(n, k)}. The C++ solution buckets the n values of gcd(n, i) and sums cnt[v] * 2^v with a big integer; the Python solution groups by divisors via Euler's totient, summing φ(d) * 2^{n/d}. Both divide the numerator by n. Big integers are required.
