## 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. 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 x, y;
int d1, d2, b;

void read() { cin >> x >> y >> d2 >> d1 >> b; }

void solve() {
    // We need the fewest moves for a tank that can only turn left (heading
    // +1 mod 6), must step forward on every turn, and may not repeat a
    // direction more than b times in a row, to travel from (0,0) facing d1 to
    // (x,y) facing d2. Coordinates reach 1e12, so the answer can be huge and
    // searching is hopeless. Instead we describe the shape of every legal walk,
    // notice it has only two real degrees of freedom, and pin those down with a
    // bit of algebra and one binary search. The plan: (1) see that a walk is
    // forced to be a chain of direction "runs"; (2) reduce its freedom to a
    // start direction and a lap count m; (3) for a fixed m turn the best run
    // lengths into a one-variable convex minimisation g(m); (4) optimise m.
    //
    // Shape of a walk. Turning is one-way and every turn also steps forward, so
    // the heading never decreases: the directions we use, in order, form a
    // contiguous arc f, f+1, f+2, ... read mod 6. We can neither skip a
    // direction (each +1 costs a move) nor return to an earlier one without
    // going the long way around, so the walk breaks into runs: one maximal run
    // per direction along that arc, each run 1..b moves long. The first move is
    // the only special case, since at the very start we either step straight
    // away (first run direction f = d1) or spend our one allowed turn first
    // (f = d1+1). Beyond that, going once around the compass is a full lap of
    // six runs. So a whole walk is captured by two numbers: the start direction
    // f (two choices) and the run count R. The final run must face d2, which
    // fixes R modulo 6, hence R = R0 + 6*m with R0 = ((d2-f) mod 6) + 1 in
    // [1..6] and m >= 0 the number of extra laps. This is the natural "short
    // opening fragment plus m identical loops" split.
    //
    // Forced vs free moves. Direction k occurs in n_k = m + c_k runs, where c_k
    // counts its runs in the base R0. We may give it any S_k in [n_k, n_k*b]
    // total moves, the answer is T = sum S_k, and the displacement must satisfy
    // sum S_k*vec_k = (x, y). Split S_k = n_k + s_k: the n_k part (one move per
    // run, forced) contributes a displacement (Cx, Cy) = sum n_k*vec_k that is
    // independent of m, because a full lap visits every direction once and each
    // direction cancels its opposite. So all the free moves s_k in [0, cap_k],
    // cap_k = (m + c_k)*(b-1), have to do is cover the fixed leftover
    // D = (x - Cx, y - Cy).
    //
    // Where the free moves go. Opposite directions cancel, so the six headings
    // act as only three axes: dir 0 and 3 on one, 1 and 4 on the next, 2 and 5
    // on the last (vec3 = -vec0, vec4 = -vec1, vec5 = -vec2). Let A = s0 - s3,
    // B = s1 - s4, C = s2 - s5 be the net signed travel along each axis.
    // Reaching a net of A on an axis costs at least |A| moves, since stepping
    // one way and back only wastes moves, so the free moves total |A|+|B|+|C|.
    //
    // The three axes are redundant in the plane, since vec2 = vec1 - vec0, so a
    // step on axis 2 equals a step on axis 1 minus one on axis 0. That leaves a
    // single real choice: how much to route through the third axis, the value
    // C. Fixing C forces the other two, because the free displacement
    // A*vec0 + B*vec1 + C*vec2 = (A - C, B + C) must equal D = (Dx, Dy), giving
    // A = Dx + C and B = Dy - C. So for a fixed lap count m the minimal number
    // of free moves is:
    //
    //     g(m) = min over feasible C of |Dx + C| + |Dy - C| + |C|.
    //
    // This is a sum of absolute values: convex in C with kinks at C = -Dx,
    // C = Dy and C = 0, so ignoring the caps its minimum sits at the middle
    // kink C* = median(-Dx, Dy, 0), which does not depend on m. The caps only
    // forbid C outside an interval I(m) that widens as m grows (more laps means
    // more budget per direction), so g(m) is that expression at C* clamped into
    // I(m).
    //
    // Optimising m. T(m) = R0 + 6*m + g(m): each extra lap adds six forced
    // moves but loosens the caps, which can only lower g, so T is convex in m.
    // We binary-search the m where its slope first turns non-negative. Running
    // all of this for both start directions f and keeping the smaller total
    // gives the answer.

    const int vx[6] = {1, 0, -1, -1, 0, 1};
    const int vy[6] = {0, 1, 1, 0, -1, -1};

    if(x == 0 && y == 0 && d1 == d2) {
        cout << 0 << '\n';
        return;
    }

    int64_t best = LLONG_MAX;
    const int64_t MHI = (int64_t)6e12;

    for(int turn = 0; turn < 2; turn++) {
        int fd = (d1 + turn) % 6;
        int R0 = (((d2 - fd) % 6 + 6) % 6) + 1;

        int64_t c[6] = {0, 0, 0, 0, 0, 0};
        for(int i = 0; i < R0; i++) {
            c[(fd + i) % 6]++;
        }

        int64_t Cx = 0, Cy = 0;
        for(int k = 0; k < 6; k++) {
            Cx += c[k] * vx[k];
            Cy += c[k] * vy[k];
        }

        int64_t Dx = x - Cx, Dy = y - Cy;
        int64_t Cstar = (-Dx) + Dy - min({-Dx, Dy, (int64_t)0}) -
                        max({-Dx, Dy, (int64_t)0});

        auto cap = [&](int64_t m, int k) -> int64_t {
            return (m + c[k]) * (b - 1);
        };

        auto eval = [&](int64_t m) -> pair<bool, int64_t> {
            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)});
            if(lo > hi) {
                return {false, 0};
            }

            int64_t C = max(lo, min(hi, Cstar));
            int64_t g = abs(Dx + C) + abs(Dy - C) + abs(C);
            return {true, R0 + 6 * m + g};
        };

        int64_t mlo = 0, mhi = MHI;
        while(mlo < mhi) {
            int64_t mid = mlo + (mhi - mlo) / 2;
            if(eval(mid).first) {
                mhi = mid;
            } else {
                mlo = mid + 1;
            }
        }

        int64_t lo = mlo, hi = MHI;
        while(lo < hi) {
            int64_t mid = lo + (hi - lo) / 2;
            if(eval(mid + 1).second >= eval(mid).second) {
                hi = mid;
            } else {
                lo = mid + 1;
            }
        }

        best = min(best, eval(lo).second);
    }

    cout << best << '\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();
        solve();
    }

    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.