<|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.

348. Twisting the Number
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



Professor Dull invented a direction in number theory. This is the theory of "twisting generators". Consider a positive integer number p. Let it contain b bits (the highest bit should be 1). Consider all possible b cycle shifts of the binary notation of the number p. Let's make a set from all this shifts and call it W(p). Note, some of the shifts can start with zero. Let's say that the number p twistingly generates the set W(p). For example, W(11) = {7, 11, 13, 14}.

The number p is called a twisting generator of the number n, if the union of all sets W(1), W(2),..., W(p) contains {1, 2,..., n} as a subset. Your task is to find the minimum generator of the given number n.

Input
The first line of the input contains the integer number n (1 ≤ n ≤ 1018).

Output
Write to the output the minimum twisting generator of the number n.

Example(s)
sample input
sample output
6
5

<|response|>
## 1) Abridged problem statement (concise)

For a positive integer \(p\) with binary length \(b\) (highest bit is 1), form the set \(W(p)\) containing all \(b\) cyclic shifts of its \(b\)-bit binary representation (shifts may start with 0).  

We call \(p\) 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** twisting generator \(p\).

---

## 2) Key observations

1. **Only numbers of the same bit-length can generate each other (for the “hard” part).**  
   Let \(L = \text{bitlen}(n)\). Any \(k < 2^{L-1}\) has fewer than \(L\) bits, and all values in \(W(k)\) also have \(< L\) bits.  
   So to generate all numbers in \([2^{L-1}, n]\), we need some \(p \ge 2^{L-1}\) (an \(L\)-bit number).

2. **Define the “best (smallest) producer” for an \(L\)-bit number.**  
   For an \(L\)-bit number \(x\), consider all \(L\)-bit cyclic rotations of \(x\). Some rotations might start with 0 (then they are not \(L\)-bit numbers), and such a number cannot appear as \(k\) in \([2^{L-1}, \dots]\).  
   Define:
   \[
   \text{min\_rotation}(x) = \min\{\text{rot}(x,k) \mid \text{rot}(x,k)\ \text{still has top bit }1\}.
   \]
   Then \(x \in W(\text{min\_rotation}(x))\).

3. **The answer equals a maximum of these minima.**  
   If \(p \ge \text{min\_rotation}(x)\), then \(k=\text{min\_rotation}(x)\) is included in \([1..p]\), so \(W(k)\) is included, hence \(x\) is generated.  
   Therefore, to cover all \(x \in [2^{L-1}, n]\), we need:
   \[
   p \ge \max_{x \in [2^{L-1}, n]} \text{min\_rotation}(x).
   \]
   This bound is also **necessary**, so:
   \[
   \boxed{\text{ans} = \max_{x \in [2^{L-1}, n]} \text{min\_rotation}(x)}.
   \]

4. **We cannot iterate to \(n\), but we only need \(O(L)\) candidates.**  
   Consider numbers \(\le n\) grouped by prefix. If we fix a prefix and allow the suffix to vary, the maximum of \(\text{min\_rotation}(x)\) in that block is attained at the largest number in the block (suffix all 1s).  
   Thus, it suffices to check:
   - \(x = n\)
   - for each bit position `pos` where \(n\) has a 1 (except the highest bit), form
     \[
     u = \text{(same higher bits as }n)\ +\ 0\ +\ \underbrace{11\ldots1}_{\text{lower bits}}
     \]
     i.e. flip that bit to 0 and set all lower bits to 1.

   There are at most \(L \le 60\) such candidates.

5. **Computing `min_rotation(x)` is \(O(L)\).**  
   Try all \(L\) cyclic shifts, keep only those with top bit still 1, and take the minimum.

Overall complexity: \(O(L^2)\) \(\approx 60^2\), easily fast enough.

---

## 3) Full solution approach

1. Read \(n\), let \(L = \text{bitlen}(n)\).
2. Define `rotate_left(x,k,L)` = cyclic left rotation within exactly \(L\) bits:
   \[
   ((x \ll k) \;|\; (x \gg (L-k))) \;\&\; ((1\ll L)-1)
   \]
3. Define `min_rotation(x,L)`:
   - iterate \(k=0..L-1\)
   - compute rotated `r`
   - accept only if the top bit (bit \(L-1\)) is 1
   - take the minimum among accepted rotations
4. Initialize `ans = min_rotation(n,L)`.
5. For each `pos` from 0 to \(L-2\):
   - if bit `pos` of \(n\) is 1, build candidate:
     - clear bits `0..pos` of `n`, keep higher bits
     - set bits `0..pos-1` to 1 (and bit `pos` becomes 0)
   - update `ans = max(ans, min_rotation(u,L))`
6. Output `ans`.

---

## 4) C++ implementation (detailed comments)

```cpp
#include <bits/stdc++.h>
using namespace std;

using int64 = long long;

// Cyclic left rotation of x by k positions within exactly 'len' bits.
// We mask to keep only 'len' bits after shifts.
static inline int64 rotate_left_fixed_len(int64 x, int k, int len) {
    int64 mask = (1LL << len) - 1;
    // k is always in [0, len-1] in our usage
    return ((x << k) | (x >> (len - k))) & mask;
}

// For an 'len'-bit number x, find the minimum cyclic rotation that
// still remains an 'len'-bit number (i.e., its highest bit stays 1).
static int64 min_rotation(int64 x, int len) {
    int64 best = x;                       // rotation by 0 is valid (x is len-bit)
    int64 topBit = 1LL << (len - 1);

    for (int k = 0; k < len; k++) {
        int64 r = rotate_left_fixed_len(x, k, len);
        if (r & topBit) {                 // valid rotation: still len-bit
            best = min(best, r);
        }
    }
    return best;
}

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

    int64 n;
    cin >> n;

    // bit length of n (n >= 1)
    int len = 64 - __builtin_clzll((unsigned long long)n);

    // Candidate set includes n itself
    int64 ans = min_rotation(n, len);

    // Enumerate "boundary" candidates:
    // for each set bit pos (except the top one), flip it to 0 and set all lower bits to 1.
    for (int pos = 0; pos + 1 < len; pos++) {
        if ((n >> pos) & 1LL) {
            // mask for low (pos+1) bits: bits 0..pos
            int64 lowMask = (1LL << (pos + 1)) - 1;

            // upper: keep bits >= pos+1, clear bits 0..pos
            int64 upper = n & ~lowMask;

            // lowerOnes: set bits 0..pos-1 to 1, keep bit pos as 0
            int64 lowerOnes = (1LL << pos) - 1;

            int64 u = upper | lowerOnes;
            ans = max(ans, min_rotation(u, len));
        }
    }

    cout << ans << "\n";
    return 0;
}
```

---

## 5) Python implementation (detailed comments)

```python
import sys

def rotate_left_fixed_len(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 cyclic rotation of x that still has the highest bit set,
    i.e. remains a 'length'-bit number.
    """
    top_bit = 1 << (length - 1)
    best = x  # x itself is a valid rotation

    for k in range(length):
        r = rotate_left_fixed_len(x, k, length)
        if r & top_bit:          # valid: still length-bit
            if r < best:
                best = r
    return best

def solve(n: int) -> int:
    length = n.bit_length()

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

    # Boundary candidates: for every set bit pos except the top bit,
    # flip it to 0 and set all lower bits to 1.
    for pos in range(0, length - 1):
        if (n >> pos) & 1:
            # Clear bits 0..pos
            upper = n & ~((1 << (pos + 1)) - 1)
            # Set bits 0..pos-1 to 1 (bit pos stays 0)
            lower_ones = (1 << pos) - 1
            u = upper | lower_ones

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

    return ans

def main() -> None:
    n = int(sys.stdin.readline().strip())
    print(solve(n))

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

These implementations follow the proven identity:
\[
\text{answer} = \max_{x \in [2^{L-1}, n]} \text{min\_rotation}(x),
\]
and compute that maximum using only \(O(L)\) carefully chosen candidates.