1. Abridged Problem Statement
Two boxes contain A and B balls respectively (A+B>0). In each move, you may transfer from one box to the other exactly as many balls as the destination box currently has. Determine the minimum number of moves to gather all balls into a single box, or report –1 if impossible.

2. Detailed Editorial
Idea of the move
- Suppose one box has x balls and the other has y balls.
- If you move from the x‐ball box into the y‐ball box, you must transfer y balls. After the move:
  • x_new = x – y
  • y_new = y + y = 2y
- If you move from y to x, you transfer x balls: results (2x, y – x).

Greedy strategy
- At each step, always transfer from the larger box into the smaller one. This keeps the numbers manageable and is forced by the rule if you want to reduce one of them to zero.

Simulation
1. Let (a,b) = (A,B).
2. Repeat up to some limit (e.g. 60 moves suffices given bounds):
   a. If a>b, swap them. Now a ≤ b.
   b. If a==0, all balls are in the other box; output moves used.
   c. Perform the move “from b into a”:
      • b := b – a
      • a := 2*a
   d. Increment step count.
3. If loop ends without a reaching 0, output –1.

Why it works and is fast
- The total number of balls T = A+B stays constant.
- Each move increases one pile while strictly decreasing the other; if termination is possible, this greedy order reaches it in O(log T) steps.
- Since T < 2^31, doubling the smaller pile more than 31 times would exceed T, so if you haven’t terminated in ~31 moves you never will. Checking up to 60 is safe and constant‐time.

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;
}

void solve() {
    // (a, b), (a > b)
    // (a - b, 2b)
    // (a - 3b, 4b)
    // (a - 7b, 8b)
    // ...

    for(int steps = 0; steps <= 60; steps++) {
        if(a > b) {
            swap(a, b);
        }

        if(a == 0) {
            cout << steps << '\n';
            return;
        }

        b -= a;
        a *= 2;
    }

    cout << -1 << '\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 with Detailed Comments
```python
import sys

def moves_to_unify(a, b):
    """
    Return the minimum number of moves to gather all balls in one box
    under the rule: transfer from one box to the other exactly as many
    balls as are currently in the destination box. Return -1 if impossible.
    """
    for steps in range(61):
        # Always make 'a' the smaller or equal box
        if a > b:
            a, b = b, a
        # If the smaller box is empty, we're done
        if a == 0:
            return steps
        # Move from b to a: b loses a, a doubles
        b -= a
        a <<= 1  # equivalent to a *= 2
    # If we reach here, no solution within 60 moves → impossible
    return -1

def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    A, B = map(int, data[:2])
    result = moves_to_unify(A, B)
    print(result)

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

5. Compressed Editorial
- Always transfer from the larger box to the smaller.
- Let (a,b) be counts with a ≤ b. One move: (a,b) → (2a, b–a).
- Repeat until a=0 (all balls in one box) or until a fixed limit (60 moves).
- Output move count if a reaches zero; otherwise –1.
