## 1. Abridged problem statement

A tank moves on a hexagonal grid. It starts at cell `(0, 0)` facing direction `d1`, and must reach cell `(x, y)` facing direction `d2`.

In one turn, it can either:

- move forward in its current direction, or
- turn left by one direction modulo `6`, then move forward.

It cannot move more than `b` consecutive times in the same direction.

Given `x, y, d2, d1, b`, find the minimum number of turns needed.

Constraints include `|x|, |y| ≤ 10^12`, so simulation or graph search is impossible.

---

## 2. Detailed editorial

### Direction system

The six hex-grid directions are represented as vectors:

```text
0: ( 1,  0)
1: ( 0,  1)
2: (-1,  1)
3: (-1,  0)
4: ( 0, -1)
5: ( 1, -1)
```

Notice:

```text
dir 3 = -dir 0
dir 4 = -dir 1
dir 5 = -dir 2
dir 2 = dir 1 - dir 0
```

---

### Shape of any valid path

The tank can only keep its current direction or increase it by `1 mod 6`.

Therefore, the sequence of directions used by the tank consists of consecutive direction blocks:

```text
f, f + 1, f + 2, ...
```

modulo `6`.

Each maximal block of equal direction is called a **run**. Each run has length between `1` and `b`.

At the first move, the tank has two choices:

1. move immediately in direction `d1`,
2. turn left first and move in direction `(d1 + 1) mod 6`.

So the first run direction `f` is either:

```text
d1
```

or

```text
d1 + 1 mod 6
```

The last run direction must be `d2`.

If the first run direction is `f`, then the number of runs `R` must satisfy:

```text
f + R - 1 ≡ d2 mod 6
```

Thus:

```text
R = R0 + 6m
```

where:

```text
R0 = ((d2 - f) mod 6) + 1
```

and `m ≥ 0` is the number of extra full cycles around the six directions.

---

### Forced moves and extra moves

Each run must have at least one move.

For a fixed starting direction `f` and lap count `m`, direction `k` appears in:

```text
n[k] = m + c[k]
```

runs, where `c[k]` is `1` if direction `k` appears in the initial fragment of `R0` runs, otherwise possibly `0`.

Each of those runs contributes at least one forced move, so total forced moves are:

```text
R0 + 6m
```

Each run can have up to `b` moves, so direction `k` can receive extra moves:

```text
s[k] ∈ [0, (m + c[k]) * (b - 1)]
```

Let:

```text
cap[k] = (m + c[k]) * (b - 1)
```

A full cycle of all six directions has zero displacement, so the displacement of the forced one-move-per-run part depends only on the initial fragment `c[k]`, not on `m`.

Let this forced displacement be `(Cx, Cy)`.

Then the extra moves must create displacement:

```text
Dx = x - Cx
Dy = y - Cy
```

---

### Reducing the six directions to one variable

Pair opposite directions:

```text
axis 0: direction 0 vs direction 3
axis 1: direction 1 vs direction 4
axis 2: direction 2 vs direction 5
```

Define signed net movement on these axes:

```text
A = s[0] - s[3]
B = s[1] - s[4]
C = s[2] - s[5]
```

Because:

```text
dir 2 = dir 1 - dir 0
```

the displacement from extras is:

```text
A * dir0 + B * dir1 + C * dir2
= (A - C, B + C)
```

We need this to equal `(Dx, Dy)`:

```text
A - C = Dx
B + C = Dy
```

So:

```text
A = Dx + C
B = Dy - C
```

Now only one variable remains: `C`.

For a fixed `C`, the minimum number of extra moves is:

```text
|A| + |B| + |C|
= |Dx + C| + |Dy - C| + |C|
```

---

### Feasibility interval for `C`

For each axis, the signed net movement must be achievable within the capacities.

For axis `0`:

```text
A = s[0] - s[3]
-cap[3] ≤ A ≤ cap[0]
```

Since `A = Dx + C`:

```text
-cap[3] - Dx ≤ C ≤ cap[0] - Dx
```

For axis `1`:

```text
B = s[1] - s[4]
-cap[4] ≤ B ≤ cap[1]
```

Since `B = Dy - C`:

```text
Dy - cap[1] ≤ C ≤ Dy + cap[4]
```

For axis `2`:

```text
C = s[2] - s[5]
-cap[5] ≤ C ≤ cap[2]
```

So the valid interval is:

```text
lo = max(-cap[5], -cap[3] - Dx, Dy - cap[1])
hi = min( cap[2],  cap[0] - Dx, Dy + cap[4])
```

If `lo > hi`, this `m` is infeasible.

---

### Minimizing extra moves for fixed `m`

For feasible `m`, minimize:

```text
g(C) = |Dx + C| + |Dy - C| + |C|
```

over:

```text
C ∈ [lo, hi]
```

This is a convex piecewise-linear function.

Without the interval restriction, its minimum is at the median of:

```text
-Dx, Dy, 0
```

Call this value `Cstar`.

With the interval restriction, the optimal `C` is:

```text
C = clamp(Cstar, lo, hi)
```

Then:

```text
answer(m) = R0 + 6m + g(C)
```

---

### Optimizing over `m`

As `m` increases, capacities only increase, so feasibility is monotonic.

First, binary search the smallest feasible `m`.

Then, because the total answer as a function of `m` is discrete convex/unimodal, binary search the first `m` where:

```text
answer(m + 1) >= answer(m)
```

Do this for both possible first directions:

```text
d1
(d1 + 1) mod 6
```

Take the smaller answer.

Special case: if the tank is already at `(0, 0)` facing `d2 == d1`, answer is `0`.

Complexity:

```text
O(log 10^12)
```

with constant memory.

---

## 3. Provided C++ solution with detailed comments

```cpp
#include <bits/stdc++.h> // Includes almost all standard C++ headers.

using namespace std; // Allows using standard library names without std:: prefix.

// Output operator for pairs.
// Not essential for this solution, but included in the original template.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second; // Print the two pair elements.
}

// Input operator for pairs.
// Not essential here, but useful in general.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second; // Read the two pair elements.
}

// Input operator for vectors.
// Reads all existing elements of a vector.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) { // Iterate over vector elements by reference.
        in >> x;      // Read each element.
    }
    return in;        // Return stream to allow chaining.
};

// Output operator for vectors.
// Prints elements separated by spaces.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {  // Iterate over vector elements.
        out << x << ' '; // Print each element followed by a space.
    }
    return out;       // Return stream to allow chaining.
};

// Target coordinates.
int64_t x, y;

// Final direction d2, initial direction d1, and max same-direction run length b.
int d1, d2, b;

// Reads input.
// Input order is x y d2 d1 b.
void read() {
    cin >> x >> y >> d2 >> d1 >> b;
}

void solve() {
    // Direction vectors for the hexagonal coordinate system.
    // Direction 0 = (1, 0), direction 1 = (0, 1), etc.
    const int vx[6] = {1, 0, -1, -1, 0, 1};
    const int vy[6] = {0, 1, 1, 0, -1, -1};

    // If the tank is already in the target cell and already has the needed
    // direction, no move is required.
    if(x == 0 && y == 0 && d1 == d2) {
        cout << 0 << '\n';
        return;
    }

    // Best answer over the two possible first-move directions.
    int64_t best = LLONG_MAX;

    // Large enough upper bound for binary searches on the number of full laps.
    const int64_t MHI = (int64_t)6e12;

    // The first move can be either:
    // turn = 0: move in initial direction d1,
    // turn = 1: turn left once, then move in direction d1 + 1.
    for(int turn = 0; turn < 2; turn++) {
        // First direction actually used.
        int fd = (d1 + turn) % 6;

        // Minimal number of runs needed so that the last run direction is d2.
        // If fd + R0 - 1 == d2 mod 6, then R0 is in [1, 6].
        int R0 = (((d2 - fd) % 6 + 6) % 6) + 1;

        // c[k] counts how many times direction k appears in the base fragment
        // of R0 consecutive runs.
        int64_t c[6] = {0, 0, 0, 0, 0, 0};

        // Fill the base fragment directions.
        for(int i = 0; i < R0; i++) {
            c[(fd + i) % 6]++;
        }

        // Compute displacement contributed by the mandatory one move
        // in each base-fragment run.
        int64_t Cx = 0, Cy = 0;

        // Full extra laps have zero net displacement, so only c[k] matters.
        for(int k = 0; k < 6; k++) {
            Cx += c[k] * vx[k];
            Cy += c[k] * vy[k];
        }

        // Remaining displacement that must be produced by extra moves.
        int64_t Dx = x - Cx;
        int64_t Dy = y - Cy;

        // The unconstrained minimizer of:
        // |Dx + C| + |Dy - C| + |C|
        // is median(-Dx, Dy, 0).
        //
        // This expression computes the median as sum - min - max.
        int64_t Cstar = (-Dx) + Dy - min({-Dx, Dy, (int64_t)0}) -
                        max({-Dx, Dy, (int64_t)0});

        // Capacity of extra moves in direction k for a given number of full
        // laps m. Direction k appears m + c[k] times as a run, and each run
        // can receive at most b - 1 extra moves.
        auto cap = [&](int64_t m, int k) -> int64_t {
            return (m + c[k]) * (b - 1);
        };

        // Evaluates whether a given m is feasible, and if so returns the
        // minimal total number of moves for this m.
        auto eval = [&](int64_t m) -> pair<bool, int64_t> {
            // Feasible interval for the signed net movement C on axis 2.
            //
            // Axis 2 condition:
            // -cap[5] <= C <= cap[2]
            //
            // Axis 0 condition:
            // -cap[3] <= Dx + C <= cap[0]
            //
            // Axis 1 condition:
            // -cap[4] <= Dy - C <= cap[1]
            int64_t lo = max({-cap(m, 5), -cap(m, 3) - Dx, Dy - cap(m, 1)});
            int64_t hi = min({cap(m, 2), cap(m, 0) - Dx, Dy + cap(m, 4)});

            // Empty interval means impossible for this m.
            if(lo > hi) {
                return {false, 0};
            }

            // Clamp the unconstrained optimum Cstar to the feasible interval.
            int64_t C = max(lo, min(hi, Cstar));

            // Minimal number of extra moves needed for this C.
            int64_t g = abs(Dx + C) + abs(Dy - C) + abs(C);

            // Total moves = forced one move per run + extra moves.
            return {true, R0 + 6 * m + g};
        };

        // First binary search:
        // find the smallest m for which the target is reachable.
        int64_t mlo = 0, mhi = MHI;

        while(mlo < mhi) {
            // Middle point.
            int64_t mid = mlo + (mhi - mlo) / 2;

            // Feasibility is monotonic: once feasible, larger m stays feasible.
            if(eval(mid).first) {
                mhi = mid;
            } else {
                mlo = mid + 1;
            }
        }

        // Second binary search:
        // the answer as a function of m is discrete convex/unimodal.
        // Find the first m where answer(m + 1) >= answer(m).
        int64_t lo = mlo, hi = MHI;

        while(lo < hi) {
            // Middle point.
            int64_t mid = lo + (hi - lo) / 2;

            // If increasing m no longer improves the answer, optimum is at
            // mid or to the left.
            if(eval(mid + 1).second >= eval(mid).second) {
                hi = mid;
            } else {
                // Otherwise answer is still decreasing, so go right.
                lo = mid + 1;
            }
        }

        // Update global best using the optimal m for this starting direction.
        best = min(best, eval(lo).second);
    }

    // Print the best answer over the two possible first directions.
    cout << best << '\n';
}

int main() {
    // Speeds up C++ input/output.
    ios_base::sync_with_stdio(false);

    // Unties cin from cout for faster input.
    cin.tie(nullptr);

    // There is only one test case.
    int T = 1;

    // Original template supports multiple tests, but this problem has one.
    // cin >> T;

    // Process the test case.
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    // Successful program termination.
    return 0;
}
```

---

## 4. Python solution with detailed comments

```python
import sys


def solve():
    # Read input.
    # Input order is: x y d2 d1 b
    data = list(map(int, sys.stdin.read().split()))
    x, y, d2, d1, b = data

    # Hex-grid direction vectors.
    vx = [1, 0, -1, -1, 0, 1]
    vy = [0, 1, 1, 0, -1, -1]

    # If already at the target position with the target heading,
    # zero turns are needed.
    if x == 0 and y == 0 and d1 == d2:
        print(0)
        return

    # Large enough upper bound for number of full laps.
    MHI = 6_000_000_000_000

    # Store the best answer over the two possible first directions.
    best = None

    # First move can be:
    # turn = 0: move in direction d1,
    # turn = 1: turn left first, then move in direction d1 + 1.
    for turn in range(2):
        # First direction used.
        fd = (d1 + turn) % 6

        # Minimal number of direction-runs needed so that the final run
        # direction is d2.
        R0 = ((d2 - fd) % 6) + 1

        # c[k] = number of times direction k appears in the base fragment
        # of R0 consecutive runs.
        c = [0] * 6
        for i in range(R0):
            c[(fd + i) % 6] += 1

        # Compute forced displacement from one mandatory move in each run
        # of the base fragment.
        #
        # Extra full laps over all 6 directions have zero displacement,
        # so they do not affect this forced displacement.
        Cx = 0
        Cy = 0
        for k in range(6):
            Cx += c[k] * vx[k]
            Cy += c[k] * vy[k]

        # Remaining displacement to be produced by extra moves.
        Dx = x - Cx
        Dy = y - Cy

        # The function to minimize over variable C is:
        # |Dx + C| + |Dy - C| + |C|.
        #
        # Its unconstrained optimum is median(-Dx, Dy, 0).
        vals = [-Dx, Dy, 0]
        vals.sort()
        Cstar = vals[1]

        def cap(m, k):
            """
            Maximum number of extra moves allowed in direction k.

            Direction k appears m + c[k] times as a run.
            Each run already has one mandatory move and can receive
            at most b - 1 extra moves.
            """
            return (m + c[k]) * (b - 1)

        def evaluate(m):
            """
            For a fixed number of full laps m:
            - determine feasibility,
            - if feasible, compute the minimal total number of moves.
            """

            # Let C be the signed net movement along axis 2:
            # C = s[2] - s[5].
            #
            # The other signed axes are forced:
            # A = Dx + C = s[0] - s[3]
            # B = Dy - C = s[1] - s[4]
            #
            # Apply capacity constraints for all three axes.

            lo = max(
                -cap(m, 5),          # from -cap[5] <= C
                -cap(m, 3) - Dx,     # from -cap[3] <= Dx + C
                Dy - cap(m, 1),      # from Dy - C <= cap[1]
            )

            hi = min(
                cap(m, 2),           # from C <= cap[2]
                cap(m, 0) - Dx,      # from Dx + C <= cap[0]
                Dy + cap(m, 4),      # from -cap[4] <= Dy - C
            )

            # No feasible value of C.
            if lo > hi:
                return False, 0

            # Clamp the unconstrained optimum to the feasible interval.
            C = max(lo, min(hi, Cstar))

            # Minimum number of extra moves.
            extra = abs(Dx + C) + abs(Dy - C) + abs(C)

            # Forced moves are one per run:
            # R0 runs in the base fragment and 6m runs from full laps.
            total = R0 + 6 * m + extra

            return True, total

        # Binary search for the smallest feasible m.
        left = 0
        right = MHI

        while left < right:
            mid = (left + right) // 2

            feasible, _ = evaluate(mid)

            # Feasibility is monotonic in m.
            if feasible:
                right = mid
            else:
                left = mid + 1

        first_feasible = left

        # Now minimize the unimodal/discrete-convex answer function.
        left = first_feasible
        right = MHI

        while left < right:
            mid = (left + right) // 2

            _, value_mid = evaluate(mid)
            _, value_next = evaluate(mid + 1)

            # If moving to mid + 1 does not improve the answer,
            # the optimum is at mid or to the left.
            if value_next >= value_mid:
                right = mid
            else:
                # Otherwise the function is still decreasing.
                left = mid + 1

        # Best value for this initial first direction.
        _, candidate = evaluate(left)

        # Update global best.
        if best is None or candidate < best:
            best = candidate

    print(best)


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

---

## 5. Compressed editorial

A valid path consists of direction runs. Since the tank can only continue straight or turn left by one, the run directions are consecutive modulo `6`.

The first run direction is either `d1` or `d1 + 1`. For a fixed first direction `f`, the number of runs must be:

```text
R = R0 + 6m
R0 = ((d2 - f) mod 6) + 1
```

where `m` is the number of extra full cycles.

Each run has length at least `1`, so one move per run is forced. Direction `k` appears `m + c[k]` times, so it can receive:

```text
cap[k] = (m + c[k]) * (b - 1)
```

extra moves.

The forced displacement from full cycles is zero, so only the initial fragment contributes. Let the remaining displacement be `(Dx, Dy)`.

Group opposite directions into three signed axes:

```text
A = s0 - s3
B = s1 - s4
C = s2 - s5
```

Since direction `2 = direction 1 - direction 0`, the extra displacement is:

```text
(A - C, B + C)
```

Thus:

```text
A = Dx + C
B = Dy - C
```

For fixed `m`, only `C` is free.

Capacity constraints give an interval:

```text
lo = max(-cap5, -cap3 - Dx, Dy - cap1)
hi = min( cap2,  cap0 - Dx, Dy + cap4)
```

If `lo > hi`, this `m` is impossible.

Otherwise minimize:

```text
|Dx + C| + |Dy - C| + |C|
```

over `[lo, hi]`.

The unconstrained optimum is:

```text
median(-Dx, Dy, 0)
```

Clamp it into `[lo, hi]`.

Then:

```text
answer(m) = R0 + 6m + extra
```

Feasibility is monotonic in `m`, so binary search the first feasible `m`. The answer function is unimodal/discrete convex, so binary search the first `m` where:

```text
answer(m + 1) >= answer(m)
```

Try both possible first directions and output the minimum.