## 1) Abridged problem statement

For a positive integer \(p\) with binary length \(b\) (highest bit is 1), consider all \(b\) cyclic left shifts of its \(b\)-bit binary representation (shifts may start with 0). Let \(W(p)\) be the set of numbers obtained from these shifts.

We say \(p\) is a *twisting generator* of \(n\) if
\[
\bigcup_{k=1}^{p} W(k) \supseteq \{1,2,\dots,n\}.
\]
Given \(1 \le n \le 10^{18}\), output the minimum such \(p\).

---

## 2) Detailed Editorial

### Key definitions

- Let `len(x)` be the number of bits in \(x\) (highest bit 1).
- For a fixed bit-length `len`, define `rot(x, k)` = cyclic left rotation of the `len`-bit representation by `k`.
- For a number \(x\) of length `len`, define:
  \[
  \text{min\_rotation}(x) = \min\{ r \mid r = \text{rot}(x,k),\ r\text{'s highest bit is 1}\}.
  \]
  (We only allow rotations that still represent a `len`-bit number, i.e. highest bit remains 1; otherwise that rotation would correspond to a shorter-length number and isn't produced by any \(p\) of length `len`.)

### Step 1: Reduce the task to the top bit-length

Let \(L = \text{len}(n)\). All numbers \(< 2^{L-1}\) have fewer than \(L\) bits.

- Any \(k < 2^{L-1}\) has length \(< L\), so all numbers in \(W(k)\) are also \(< 2^{L-1}\).
- Therefore, to cover numbers in \([2^{L-1}, n]\), we must use some generator \(p\) with \(L\) bits; in particular \(p \ge 2^{L-1}\).

Also, if we manage to generate every number in \([2^{L-1}, n]\), then automatically \(\{1,\dots,2^{L-1}-1\}\) is already covered by \(\bigcup_{k=1}^{2^{L-1}-1} W(k)\) (since those \(k\) themselves are included). So the "hard part" is covering the \(L\)-bit interval.

Hence the answer is some \(p \in [2^{L-1}, n]\) that ensures all numbers in \([2^{L-1}, n]\) appear in \(\bigcup_{k=1}^{p} W(k)\).

### Step 2: The crucial characterization

Consider an \(L\)-bit number \(x\). If \(p\) is at least one of \(x\)'s valid rotations, then \(W(p)\) contains \(x\) only if \(p\) itself is some rotation of \(x\). More importantly:

- Let \(m = \text{min\_rotation}(x)\). Then \(m\) is an \(L\)-bit number (highest bit 1) and \(x \in W(m)\) because \(x\) is a rotation of \(m\).
- Therefore, if our generator \(p \ge m\), then since we include all \(k \le p\), we include \(k=m\), hence \(W(m)\), hence \(x\) is generated.

So to ensure all \(x \in [2^{L-1}, n]\) are generated, it is sufficient that
\[
p \ge \max_{x \in [2^{L-1}, n]} \text{min\_rotation}(x).
\]

It's also necessary: if \(p\) is smaller than that maximum, pick an \(x\) achieving it; then \(m=\text{min\_rotation}(x) > p\), so \(k=m\) is not included and there is no smaller \(k\) (of length \(L\)) whose rotations include \(x\) (by definition of minimum), so \(x\) cannot be generated.

Thus the answer is exactly:
\[
\boxed{\text{ans} = \max_{x \in [2^{L-1}, n]} \text{min\_rotation}(x)}.
\]

### Step 3: How to compute it fast (avoid iterating up to \(n\))

Naively iterating all \(x\) up to \(10^{18}\) is impossible. The solution uses a monotonic/"prefix" observation.

Write \(n\) in binary (length \(L\)). Consider fixing a prefix and then decreasing \(n\) by turning some 1-bit into 0, and setting all less significant bits to 1 (to remain as large as possible under \(n\)).

For each bit position `pos` where \(n\) has a 1, define:
- Keep all bits above `pos` unchanged.
- Set bit `pos` to 0.
- Set all lower bits (`pos-1...0`) to 1.

This gives a number:
\[
u = (n \text{ with bits } \le pos \text{ replaced by } 0\,111\ldots111).
\]
This \(u\) is the maximum number below \(n\) that shares the same higher prefix but has a smaller value due to flipping that one bit.

The key claim used by the code:
- Over any such "block" of numbers sharing a prefix, the maximum value of `min_rotation(x)` occurs at the block's maximum element (i.e., with the suffix all 1s).
- Therefore, it is enough to evaluate `min_rotation` only on:
  - \(n\) itself
  - each such \(u\) derived from flipping a 1-bit of \(n\) and filling the suffix with 1s

There are only \(O(L)\) such candidates (at most 60 for \(10^{18}\)).

For each candidate, `min_rotation` is computed by trying all \(L\) rotations, each in O(1), so O(L) per candidate.

Total complexity: \(O(L^2)\) ≈ \(60^2\), easily within limits.

### Step 4: Computing `min_rotation`

For each shift \(k = 0..L-1\):
- Compute cyclic left rotation `r`.
- Keep it only if `r` still has the top bit set (i.e. is an \(L\)-bit number).
- Track the minimum such `r`.

Bit rotation for length `L`:
\[
\text{rot}(x,k) = ((x \ll k)\ |\ (x \gg (L-k)))\ \&\ ((1 \ll L)-1)
\]

---

## 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;
};

int64_t n;

void read() { cin >> n; }

int64_t rotate_left(int64_t x, int k, int len) {
    return ((x << k) | (x >> (len - k))) & ((1LL << len) - 1);
}

int64_t min_rotation(int64_t x, int len) {
    int64_t res = x;
    for(int i = 0; i < len; i++) {
        int64_t r = rotate_left(x, i, len);
        if((r >> (len - 1)) & 1) {
            res = min(res, r);
        }
    }
    return res;
}

void solve() {
    // It's clear that n is an upper bound on the answer, so now we should check
    // if something lower is satisfactory. It's also clear that the answer
    // should have the same number of bits as n as the problem statement
    // requires applying the operation to numbers with highest bit set.
    // Therefore, we are interested in the smallest number from [2^(len-1);n]
    // that could generate all of [2^(len-1);n]. Let's try to get a lower bound
    // for the answer. One candidate is to look at each number from [1;n] and
    // find the smallest shift it can be derived from. Or the lower bound would
    // be:
    //
    //     ans = max(min_rotation(i) for i in [2^(len-1);n])
    //
    // Which after some thinking, we can notice is actually the answer to the
    // problem. We ans >= min_rotation(i) for all i in [2^(len-1);n] by
    // definition, so W(min_rotation(i)) will be generated and so all
    // i in [2^(len-1);n] will also be generated.
    //
    // Now that we know how to find the answer is the lower bound, we should try
    // to find it quickly. The main observation is that for any prefix of n,
    // there is at most one sufficient candidate we can check. Say that
    // n = dec(pref || 1 || suff). Then we have the "i" values of the form
    // dec(pref || 0 || x) for any x. Out of them, it's not hard to notice that
    // min_rotation(dec(pref || 0 || 1...1)) is the largest, so we don't really
    // need to look at any of the others. Therefore, there are only O(log n)
    // different numbers to consider, and find the minimal rotation in O(log n)
    // for each, giving an overall O(log^2 n) solution.

    int len = 64 - __builtin_clzll(n);
    int64_t ans = min_rotation(n, len);

    for(int pos = 0; pos + 1 < len; pos++) {
        if((n >> pos) & 1) {
            int64_t u = (n & ~((1LL << (pos + 1)) - 1)) | ((1LL << pos) - 1);
            ans = max(ans, min_rotation(u, len));
        }
    }

    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();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys

def rotate_left(x: int, k: int, length: int) -> int:
    """
    Cyclic left-rotate x by k within exactly 'length' bits.
    """
    mask = (1 << length) - 1
    k %= length
    return ((x << k) | (x >> (length - k))) & mask

def min_rotation(x: int, length: int) -> int:
    """
    Minimum rotation of x among rotations that keep the top bit set
    (i.e., result is still a 'length'-bit number).
    """
    best = x
    top_bit = 1 << (length - 1)
    for k in range(length):
        r = rotate_left(x, k, length)
        if r & top_bit:          # valid only if highest bit remains 1
            if r < best:
                best = r
    return best

def solve(n: int) -> int:
    # Bit-length of n (n>=1)
    length = n.bit_length()

    # Start with candidate x = n
    ans = min_rotation(n, length)

    # Enumerate boundary numbers u by flipping each 1-bit (except top bit) to 0
    # and making all lower bits 1.
    # pos = 0..length-2
    for pos in range(0, length - 1):
        if (n >> pos) & 1:
            # Clear bits 0..pos in n, keep higher bits
            upper = n & ~((1 << (pos + 1)) - 1)
            # Set lower bits 0..pos-1 to 1, keep bit pos as 0
            lower_ones = (1 << pos) - 1
            u = upper | lower_ones

            ans = max(ans, min_rotation(u, length))

    return ans

def main():
    n_str = sys.stdin.readline().strip()
    n = int(n_str)
    print(solve(n))

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

---

## 5) Compressed editorial

Let \(L=\text{bitlen}(n)\). Only \(L\)-bit numbers can generate \(L\)-bit targets, so the answer lies in \([2^{L-1}, n]\). For any \(L\)-bit \(x\), define \(\text{min\_rotation}(x)\) as the smallest cyclic rotation of \(x\) that still has the top bit 1. Then \(x \in W(\text{min\_rotation}(x))\), so \(x\) is generated once \(p \ge \text{min\_rotation}(x)\). Hence the minimal generator is:
\[
\text{ans}=\max_{x\in[2^{L-1},n]} \text{min\_rotation}(x).
\]
To avoid iterating all \(x\), it suffices to evaluate this maximum on only \(O(L)\) "boundary" values: \(n\) and, for each set bit position \(pos<L-1\), the number obtained by flipping bit \(pos\) of \(n\) from 1 to 0 and setting all lower bits to 1. Compute `min_rotation` by trying all \(L\) rotations (masking to \(L\) bits and keeping only rotations with top bit 1). Total \(O(L^2)\) (~3600 ops).
