1. **Abridged problem statement**

You start with an integer `A` (2 ≤ A < B ≤ 10¹²).
In one move, if the current number is `x`, you must choose a divisor `d` of `x` such that `1 < d < x`, and replace `x` with `x + d`.

You want to reach `B` from `A` by repeating this operation.

Output:
- `"Impossible"` if there is no sequence of moves.
- Otherwise, output all numbers in the sequence from `A` to `B` inclusive, one per line, with at most 500 numbers in total (any valid sequence ≤ 500 is accepted).


---

2. **Detailed editorial**

### 2.1. Observations

**Move rule:**
From `x` you can go to `x + d` where `d` is a proper divisor: `1 < d < x` and `d | x`.

- If `x` is **prime**, then it has no divisors strictly between 1 and itself.
  So from a prime `x` you cannot move at all.
- Any **odd** number has only **odd** divisors. So adding such a divisor preserves parity:
  - odd + odd = even
  - even + even = even
So:
- From **odd** you always move to **even** (since divisor must be >1 and <x, hence odd, sum becomes even).
- From **even** you always move to **even** (sum of two evens).
Thus:
- After at most one step, you reach an **even** number.
- Once you are even, you remain **even** forever.

So any path except the first step is entirely in even numbers.

This leads to a natural decomposition:
1. Handle parity / primality at the endpoints `A` and `B`.
2. Then solve the problem for **even A and even B**.

---

### 2.2. Smallest nontrivial divisor

Define:

```text
int64_t min_divisor(int64_t x) {
    for (int64_t d = 2; d * d <= x; d++) {
        if (x % d == 0) return d;
    }
    return -1; // means x is prime
}
```

- If `min_divisor(x) == -1`, then `x` is prime → no moves possible from `x`.

We will use `min_divisor` to make an odd number even with one step:
- If `A` is odd and composite: we can do `A → A + min_divisor(A)` (this will be even).
- If `B` is odd and composite: we might want to step **backwards in construction** by considering the last move to be `B - min_divisor(B) → B`.

But from the problem's perspective, we need a **forward** sequence from `A` to `B`. We can think like this:

- We will build a sequence:
  - Possibly one step from odd `A` to an even `adjusted_A`.
  - Then several steps between even numbers, from `adjusted_A` to `adjusted_B`.
  - Possibly one final step from an even predecessor to odd `B`.

We precompute:

```text
d1 = min_divisor(A)
dk = min_divisor(B)
adjusted_A = (A is odd ? A + d1 : A)
adjusted_B = (B is odd ? B - dk : B)
```

Why `adjusted_B = B - dk`?

- If `B` is odd and composite, a plausible **last step** could be `adjusted_B → B`, with move `+dk` (since `dk` divides `B` and `1 < dk < B`).
- In that last step, the number before `B` must be `B - dk`.

So we solve the even-equal problem for `adjusted_A` and `adjusted_B` and then sandwich with `A` and `B` if they differ.

**Failure conditions:**

- If `A` is odd and `d1 == -1` → `A` is prime, and no first step possible.
- If `B` is odd and `dk == -1` → no valid last step can end exactly at `B`.
- If `adjusted_A > adjusted_B` → even after making `A` larger and `B` smaller, the interval is empty, so no sequence.

In any of those cases, output `"Impossible"`.

Note: if `A` is even but `d1 == -1`, that never happens since any even number ≥ 4 has `2` as divisor. Similarly, an odd composite has some divisor, only primes give `-1`.

---

### 2.3. Even-to-even core: `solve_even(A, B)`

Now assume:

- `A` and `B` are both even.
- We must generate a sequence of valid moves from A to B.

The helper function `solve_even` implements the strategy. Its invariant:

- Keep `A` even.
- At each step choose a power of two `2^K` as the added divisor `d`.

Why is a power of two always a valid divisor?

We need to ensure:
1. `2^K` divides the current `A` (i.e., `A % 2^K == 0`).
2. `1 < 2^K < A` (proper divisor limitation).
3. And we must not overshoot `B`, so `A + 2^K <= B`.

If we can always choose such a `2^K`, and each step increases `A`, we will eventually reach `B`.

The algorithm's idea:

- Use an increasing power of two while:
  - It remains a proper divisor of `A`.
  - The next step does not overshoot `B`.

- If the current `2^K` is too big to add (would overshoot `B`), halve it until it fits (`2^K <= B - A`).

The key property:
When both `A` and `B` are even, we can always perform a sequence of such steps in at most O(log B) moves, by a reasoning about binary representation of `(B - A)` and low bits of `A`.

#### 2.3.1. Detailed reasoning from the comments

Let `A < B` be even.

We claim there exists some `K ≥ 1` such that:
- `A % 2^K == 0` (2^K divides A),
- `A + 2^K <= B`,
- and `2^K < A` (so 2^K is a proper divisor: >1 and <A).

Among all such K, choose the **largest**.

After choosing this K and performing `A ← A + 2^K`, two possibilities:

1. **K is limited by divisibility:**
   - That is, `2^(K+1)` does not divide original `A`.
   - Then `A % 2^(K+1) = 2^K`.
   - After adding `2^K`, the new `A` satisfies `A % 2^(K+1) = 0`, so `2^(K+1)` divides new `A`.
   - Therefore next step we can pick at least exponent `K+1` (i.e., a larger power of two is now available).
   - Intuitively, we are "shifting" the 1-bit at position K into a higher divisibility.

   This case can occur at most O(log A) times, because K strictly increases each time.

2. **K is limited by `A + 2^K <= B`:**
   - That means divisibility is not the bottleneck; rather we're approaching B and cannot add bigger powers without overshooting.
   - Consider the difference `D = B - A` when we first hit this case.
   - Each step we choose the largest power of two that:
     - divides current `A`, and
     - fits within remaining distance to `B`.
   - Due to divisibility properties, we can think of these steps as removing one bit at a time from the binary representation of `D`, using powers of two.
   - Because `D` is even, and divisibility for smaller powers is retained, we can resolve the remaining difference in fewer than K steps.

Combining both cases:
- At most O(log B) "case 1" steps to grow the usable power of two.
- Then at most O(log B) "case 2" steps to consume the remaining difference.
- Total < 2 log₂(B) < 500 for all B up to 10¹² (log₂10¹² ≈ 40).

Thus the method respects the ≤500 constraint.

---

### 2.4. Implementation of `solve_even`

```text
vector<int64_t> solve_even(int64_t A, int64_t B) {
    assert(A % 2 == 0 && B % 2 == 0);

    int64_t pw2 = 2;          // current power of 2 we try to use
    vector<int64_t> steps;
    while (A < B) {
        // Increase pw2 while:
        //  1) adding 2*pw2 still doesn't overshoot B: A + 2*pw2 <= B
        //  2) 2*pw2 is still a proper divisor candidate: 2*pw2 < A
        //  3) 2*pw2 divides A: A % (2*pw2) == 0
        while (A + 2 * pw2 <= B && 2 * pw2 < A && A % (2 * pw2) == 0) {
            pw2 *= 2;
        }

        // If pw2 is too large to add (A + pw2 > B), shrink it
        while (A + pw2 > B) {
            pw2 /= 2;
        }

        steps.push_back(A);  // record the current A
        A += pw2;            // make a move: A -> A + pw2
    }

    steps.push_back(B);      // final B
    return steps;
}
```

This always returns a sequence of even numbers from `A` to `B`, using only powers of two as the added divisors. Each intermediate step satisfies the rules.

---

### 2.5. Putting it all together: `solve()`

1. Compute `d1 = min_divisor(A)`, `dk = min_divisor(B)`.
2. Compute `adjusted_A` and `adjusted_B`:
   - If `A` is odd: `adjusted_A = A + d1`; else `adjusted_A = A`.
   - If `B` is odd: `adjusted_B = B - dk`; else `adjusted_B = B`.
3. Check for impossible:
   - `d1 == -1` and `A` odd: no first move.
   - `dk == -1` and `B` odd: no last move into `B`.
   - `adjusted_A > adjusted_B`: impossible to bridge.
4. Otherwise:
   - Run `solve_even(adjusted_A, adjusted_B)` → vector of numbers already including endpoints.
   - Print:
     - If `adjusted_A != A`, first print `A` (original starting number).
     - Print every number in `ans` (from `adjusted_A` to `adjusted_B`).
     - If `adjusted_B != B`, finally print `B`.

Note: `solve_even` already prints the starting and ending even numbers. The odd endpoints are handled separately, before and after.

This yields a valid sequence of moves from original `A` to `B`, respecting the rules, in ≤ 500 steps.

---

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 A, B;

void read() { cin >> A >> B; }

vector<int64_t> solve_even(int64_t A, int64_t B) {
    assert(A % 2 == 0 && B % 2 == 0);

    int64_t pw2 = 2;
    vector<int64_t> steps;
    while(A < B) {
        while(A + 2 * pw2 <= B && 2 * pw2 < A && A % (2 * pw2) == 0) {
            pw2 *= 2;
        }

        while(A + pw2 > B) {
            pw2 /= 2;
        }

        steps.push_back(A);
        A += pw2;
    }

    steps.push_back(B);
    return steps;
}

int64_t min_divisor(int64_t x) {
    for(int64_t d = 2; d * d <= x; d++) {
        if(x % d == 0) {
            return d;
        }
    }

    return -1;
}

void solve() {
    // The first thing we notice is that we don't want the shortest solution,
    // but just any with <= 500 steps. The constraints for A and B are fairly
    // large at 10^12, so it's unlikely that we can get something directly based
    // on the divisors so that it fits within 500 steps. A common idea in
    // constructive problems like this is to explore powers of 2 and then
    // achieve something that is logarithmic. In particular, let's consider the
    // case where both A < B are even. As A and B are both even, there will be
    // some K >= 1, such that A mod 2^K = 0 and A + 2^K <= B (and 2^K < A due to
    // 1 < d < A). Let us choose the largest such K and make the operation A +=
    // 2^K. There are two cases:
    //
    //     1) The constraint was that K was the largest power of 2. If that is
    //        the case, we know that A mod 2^(K+1) = 2^K. Then after the
    //        operation, A mod 2^(K+1) = 0. This means on the next step we would
    //        be able to choose K + 1.
    //
    //     2) The second constraint was bounding or A+2^K <= B. If this is the
    //        case, it's not hard to observe that we will be done in < K steps
    //        using the same procedure. This is because the difference B-K is
    //        even and the operation represents removing the largest bit from
    //        the binary notation. Note that even after adding 2^K, the
    //        divisibility by all K' < K is still kept which allows us to
    //        perform this algorithm if we always choose the largest K.
    //
    // We can have < log(B) times that we hit case 1, followed by again less
    // than log(B) times of case 2. This is logarithmic and within the 500
    // bound. Hence, we now have a solution for the case of both A and B being
    // even.
    //
    // What about the case when A or B is odd? In this case, all divisors of
    // both A and B are odd. We can notice that we can always get from all even
    // numbers to all other even numbers (apart from corner case of A = 2), so
    // it might make sense to make both A and B even. To do this, let's denote
    // by min_divisor(x) the smallest number d such that x mod d = 0 and 1 < d <
    // x. If A is odd, we will have the first operation be A += min_divisor(A).
    // If B is odd, we will have the last operation be A += min_divisor(B).
    // Clearly, it would be impossible if adjusted_A > adjusted_B, where
    // adjusted_A = A + min_divisor(A), and adjusted_B = B - min_divisor(B).
    //
    // Computing min_divisor(x) can be trivially done in O(sqrt(x)), while
    // the rest of the algorithm has O(log(B)) steps, where each step of finding
    // K can be trivially implemented in O(log(B)) too, although if we reuse the
    // last steps value of K as a start and move it as a pointer, the actual
    // amortized complexity will be just O(log(B)). Either way, this is
    // dominated by the min_divisor(x) computation.

    int64_t d1 = min_divisor(A), dk = min_divisor(B);
    int64_t adjusted_A = A, adjusted_B = B;
    if(A % 2 == 1) {
        adjusted_A = A + d1;
    }
    if(B % 2 == 1) {
        adjusted_B = B - dk;
    }
    if(d1 == -1 || dk == -1 || adjusted_A > adjusted_B) {
        cout << "Impossible" << endl;
        return;
    }

    vector<int64_t> ans = solve_even(adjusted_A, adjusted_B);
    if(adjusted_A != A) {
        cout << A << endl;
    }
    for(int64_t x: ans) {
        cout << x << endl;
    }
    if(adjusted_B != B) {
        cout << B << 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();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

4. **Python solution with detailed comments**

```python
import sys
from math import isqrt

# Read A and B from standard input
data = sys.stdin.read().strip().split()
A = int(data[0])
B = int(data[1])

def min_divisor(x: int) -> int:
    """
    Return the smallest non-trivial divisor of x (2 <= d < x).
    If x is prime, return -1.
    """
    # Try all integers d from 2 up to sqrt(x)
    # If any divides x, it's the smallest non-trivial divisor found.
    limit = isqrt(x)
    for d in range(2, limit + 1):
        if x % d == 0:
            return d
    # If we reached here, x has no non-trivial divisor: x is prime
    return -1

def solve_even(A: int, B: int):
    """
    Solve the problem for even A and even B.
    Return a list of integers representing a valid sequence
    from A to B (inclusive) under the rules.
    """
    assert A % 2 == 0 and B % 2 == 0

    pw2 = 2              # current power of two to try
    steps = []           # store the path from A to B

    # Continue while we haven't reached B
    while A < B:
        # Try doubling pw2 as long as:
        #  1) A + 2*pw2 <= B   (does not overshoot B)
        #  2) 2*pw2 < A        (remains a proper divisor)
        #  3) A % (2*pw2) == 0 (2*pw2 divides A)
        while A + 2 * pw2 <= B and 2 * pw2 < A and A % (2 * pw2) == 0:
            pw2 *= 2      # increase to the next power of two

        # Now ensure pw2 is not too big to add.
        # While A + pw2 > B, shrink pw2 by halving it.
        while A + pw2 > B:
            pw2 //= 2

        # Record current A as part of the sequence
        steps.append(A)
        # Perform the move A -> A + pw2
        A += pw2

    # Finally include B itself
    steps.append(B)
    return steps

def main():
    global A, B

    # Compute smallest non-trivial divisors of A and B
    d1 = min_divisor(A)
    dk = min_divisor(B)

    # Start with adjusted values equal to the originals
    adjusted_A = A
    adjusted_B = B

    # If A is odd, we want to first do A -> A + d1
    if A % 2 == 1:
        adjusted_A = A + d1

    # If B is odd, we want the last move to be
    # adjusted_B -> adjusted_B + dk = B, so adjusted_B = B - dk
    if B % 2 == 1:
        adjusted_B = B - dk

    # Check for impossibilities:
    # - If A is odd prime: d1 == -1 and we can't take the first step
    # - If B is odd prime: dk == -1 and we can't end at B
    # - If adjusted_A > adjusted_B: even interval can't be traversed
    if d1 == -1 or dk == -1 or adjusted_A > adjusted_B:
        print("Impossible")
        return

    # Now adjusted_A and adjusted_B are even and adjusted_A <= adjusted_B
    seq_even = solve_even(adjusted_A, adjusted_B)

    out_lines = []

    # If A was modified (odd), print original A first
    if adjusted_A != A:
        out_lines.append(str(A))

    # Then print the even sequence from adjusted_A to adjusted_B
    for x in seq_even:
        out_lines.append(str(x))

    # If B was modified (odd), finally print B
    if adjusted_B != B:
        out_lines.append(str(B))

    # Output result
    sys.stdout.write("\n".join(out_lines) + "\n")

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

This Python program faithfully implements the same logic as the C++ solution and respects the problem's constraints.

---

5. **Compressed editorial**

We must transform `A` into `B` by repeatedly adding a proper divisor of the current number. Proper divisor means `1 < d < x` and `d | x`.

Key parity observation:
- Divisors of odd numbers are odd → from odd you go to even (odd + odd = even).
- From even, all divisors are even → you stay even (even + even = even).
So the path is at most: odd → even → even → ... → even → maybe odd, but under the rules, once you're even you never become odd again. This means:
- If `B` is odd and there is a valid sequence, the last step must be from some even `X = B - d`, where `d` is a proper divisor of `B`.

We define `min_divisor(x)` as the smallest non-trivial divisor of `x`, or `-1` if `x` is prime (no moves from a prime). Set:

- `adjusted_A = (A is odd ? A + min_divisor(A) : A)`
- `adjusted_B = (B is odd ? B - min_divisor(B) : B)`

Each adjustment is exactly one valid step (if divisor exists). If:
- `min_divisor(A) == -1` and A is odd → cannot move from A.
- `min_divisor(B) == -1` and B is odd → cannot end at B.
- Or `adjusted_A > adjusted_B` → no way to bridge between them.

Then the answer is `"Impossible"`.

Otherwise `adjusted_A` and `adjusted_B` are even and `adjusted_A ≤ adjusted_B`. Now we only need to solve the even-to-even problem.

For even numbers, we use powers of two as divisors: `d = 2^K`. We maintain a current power `pw2` and repeatedly:
- Try to increase `pw2` while `2*pw2` divides the current `A`, is less than `A`, and `A + 2*pw2 ≤ B`. This finds the largest power of two that is a valid move and doesn't overshoot.
- If current `pw2` overshoots (`A + pw2 > B`), halve it until it fits.

At each step we record `A`, then do `A += pw2`. We continue until `A == B`. This sequence is valid because each `pw2` divides `A` and is a proper divisor. Analysis shows this uses at most `O(log B)` steps: we either increase the maximum usable power of two due to better divisibility, or we are in the phase where the difference `(B - A)` is consumed by successively removing its highest power-of-two contributions. Total steps < 2 log₂ B < 500 for B ≤ 10¹².

The final sequence is:
- Optional original `A` (if odd),
- The even sequence from `adjusted_A` to `adjusted_B`,
- Optional final `B` (if odd).

This meets the problem's sequence-length guarantee and either prints `"Impossible"` or a valid sequence.
