1. **Abridged problem statement**

We have an \(N \times N\) chessboard and three kings:

- Black king at \((x_1, y_1)\)
- White king at \((x_2, y_2)\)
- Invisible “black‑white” king at \((x_3, y_3)\)

Move order in every step: **white**, then **black**, then **black‑white**.  
Each king moves like a regular chess king: to any of the 8 neighboring cells (or can stay still in effect by choosing a path, but in this problem they always follow shortest paths).

The black and white kings choose some shortest paths so that they meet *face‑to‑face*: they end up on two neighboring cells. The black‑white king also moves optimally along a non‑detour path (no zigzags that could be shortened).

The black‑white king **wins** if it can ever move onto a square currently occupied by one of the other two kings (i.e., it “attacks” that king) **before** black and white end up on neighboring cells. If one of black/white moves onto the black‑white’s cell, no capture happens.

We must determine:

- If the black‑white king has a nonzero chance to win (i.e., at least one consistent choice of shortest paths for black & white and a non‑detour path for black‑white where it intercepts).
  - Output:  
    `YES`  
    and the **minimum number of black‑white moves** needed for a possible interception.
- Otherwise:
  - Output:  
    `NO`  
    and the **minimum total number of moves of black + white** needed until they become neighbors (under optimal shortest paths).


---

2. **Detailed editorial**

### 2.1. Geometry and timing

Distance between two kings (using king moves) is the **Chebyshev distance**:

\[
D = \max(|x_1 - x_2|, |y_1 - y_2|)
\]

This is the minimum number of moves **each** of black and white must make to **reach the same square**.  
But they only need to become **neighbors**, so effectively they will meet after `D - 1` moves each (they can stop one step apart).

Total moves of both together until they are neighbors:
\[
\text{total} = 2(D - 1)
\]
But the problem asks just for “minimal total number of moves of black and white kings necessary to meet”. Because they move one after another, that is just `D-1` moves of each king; the solution prints `D-1`.

Time is discrete with move order:

1. White moves.
2. Black moves.
3. Black‑white moves.

We only care about the position of kings at **the moment when black‑white moves**, because:

- An interception happens only when **black‑white** moves onto the cell of some king.
- If black or white moves onto black‑white, nothing happens.

So we consider “time steps” indexed by **t** = number of moves made by the black‑white king so far.  
At each such step \(t\):

- White has made \(t\) moves.
- Black has made \(t\) moves.
- Black‑white is about to make his \(t\)-th move.

So all three make the same number of moves up to that point.

### 2.2. Black & white’s motion structure

Two ordinary kings move along some shortest paths. For any two positions \((x_1, y_1)\), \((x_2, y_2)\):

- Minimal meeting time if they wanted to share the same square is \(D = \max(dx, dy)\) where \(dx = |x_1 - x_2|\), \(dy = |y_1 - y_2|\).
- On a shortest path, in each move they reduce either \(|x_1 - x_2|\), or \(|y_1 - y_2|\), or both, and they never “go back”.

However, the problem is tricky because:

- They don’t aim to reach each other’s initial squares; they just aim to end up neighboring.
- There are many possible shortest paths.

We don’t want to enumerate all paths. Instead we want to characterize **what squares are reachable** at step \(t\) by each king, consistent with moving along some shortest path to meet the other.

The official solution idea (which this code implements) can be described like this:

1. Assume w.l.o.g. that the **x‑axis difference is at least as big** as the y‑axis difference:  
   If \(|x_1 - x_2| < |y_1 - y_2|\), swap x and y for all three kings to ensure \(dx \ge dy\).  
   This simplifies direction / range considerations.

2. Let:
   - \(D = \max(dx, dy) = dx\).
   - They can become neighbors roughly after \(D-1\) moves each.
   - Because black‑white moves once per “pair” of moves (one white + one black), if we think in “paired steps” then the number of black‑white moves before that is about \(D/2\). The code uses
     \[
     \text{max\_steps} = \frac{D}{2} - 1
     \]
   - If `max_steps <= 0`, then black‑white has **no time** to intersect **before** black and white are already neighbors (we’ll justify later).

So we only need to check times \(t = 1, 2, ..., \text{max\_steps}\). If interception hasn’t happened by then, it can never happen before black & white meet.

### 2.3. Black‑white king’s reachable region

The black‑white king starts at \(c = (x_3, y_3)\).

It also moves along a shortest (non‑detour) path; that simply means it always reduces its Chebyshev distance to some final target, and in t moves it cannot be outside the t‑radius king metric.

So after \(t\) moves, black‑white can be in any cell within Chebyshev distance \(t\) of \((x_3, y_3)\):

\[
x_3 - t \le x \le x_3 + t \\
y_3 - t \le y \le y_3 + t
\]

This is an axis‑aligned square (in Chebyshev metric) centered at `c` with radius `t`. In the code:

```
int x3 = c.first - t, x4 = c.first + t;
int y3 = c.second - t, y4 = c.second + t;
```

So black‑white’s possible positions at time `t` form the rectangle `[x3, x4] × [y3, y4]`.

### 2.4. Ranges of black & white kings at time t

We want to know: at time \(t\), what cells can black or white be on if they are:

- Moving along some shortest path (length `D`) to meet the other king.
- With the constraint that their final distance becomes 1 at time `D-1`.

The core simplification the author uses:

- After we ensure \(dx \ge dy\), the x‑coordinate difference controls the meeting.
- Each of the two kings essentially moves along a straight line in x (direction either +1 or −1) on their shortest paths.

Let’s define:

- `x_dir` is +1 if black’s x is to the right of white’s x, otherwise −1:
  
  ```
  int x_dir = (b.first > a.first) ? 1 : -1;
  ```

We only check **interception with paths where at each step they move roughly straight in x**. That’s correct because the problem only needs to know if **some** shortest path choice leads to interception: if interception is possible at all, we can assume this kind of “monotone” path in x.

After `t` moves, a king starting at `(kx, ky)` and moving in x with direction `dir` will have x-coordinate:

\[
x = kx + dir \cdot t
\]

The y coordinate can vary, subject to:

- King can move at most `t` steps from its original y: `ky - t` to `ky + t`, clamped by 1..N (board bounds).
- Also, it must stay on some shortest path to the other king.

The “shortest path to other king” y‑constraint is encoded as:

- The initial vertical distance is `dy`.
- Over the entire D steps, they can only move so that the sum of absolute differences is consistent with decreasing the Chebyshev distance.
- Effectively, at time `t` the y-coordinate must be within some strip depending on the opponent’s position and `D`.

In code, for king `king` and opponent `other`:

```
int y_min =
    max(clamp(king.second - t, 1, n),
        clamp(other.second - D + t, 1, n));
int y_max =
    min(clamp(king.second + t, 1, n),
        clamp(other.second + D - t, 1, n));
```

Intuition:

- `king.second ± t` is the natural t-step vertical reach.
- `other.second - D + t` and `other.second + D - t` come from the fact that along a shortest path of length `D`, the vertical distance between king and other evolves within certain bounds; by time `t` the king cannot be too far above or below the opponent if it still wants to end at distance at most 1 when time reaches `D-1`.

So at step t, each king’s **possible y values** are in `[y_min, y_max]` with a **fixed x** as above.

### 2.5. Checking intersection at time t

Given a candidate king at time `t`:

- Its x is fixed: `x = king.first + dir * t`
- Its y can be anywhere in `[y_min, y_max]`.

We must determine whether **any** cell in its reachable vertical segment intersects the black‑white king’s t-radius square `[x3,x4] × [y3,y4]`.

The code function:

```
auto check = [&](pair<int, int> king, pair<int, int> other, int dir) {
    int x = king.first + dir * t;
    int y_min = ...;
    int y_max = ...;

    if(x < x3 || x > x4) {
        return false;
    }
    if(x == x3 || x == x4) {
        return max(y_min, y3) <= min(y_max, y4);
    }
    return (y_min <= y3 && y3 <= y_max) || (y_min <= y4 && y4 <= y_max);
};
```

Interpretation:

1. If the king’s x at time t lies **outside** the black‑white square’s x span (`x3..x4`), no intersection is possible.

2. If `x` is exactly at the left or right boundary of the square (`x == x3 || x == x4`):
   - Intersection exists if the king’s vertical segment `[y_min, y_max]` overlaps the square’s vertical range `[y3, y4]`.

3. If `x` is strictly inside `(x3, x4)`:
   - Then for an intersection it suffices that there is a y in `[y_min, y_max]` that’s within `[y3,y4]`.  
   - Since black‑white controls a full vertical interval in that x‑range, we just need to see if either `y3` or `y4` lies in the king’s segment:  
     `(y_min <= y3 && y3 <= y_max) || (y_min <= y4 && y4 <= y_max)`.
   - This is equivalent to standard segment intersection but simplified under the assumption of monotonic movement.

We run this `check` for **both kings**, because black‑white could intercept either:

- First: black moving toward white:
  ```
  check(a, b, x_dir)   // a = black, b = white (or vice versa, depending on names)
  ```
- Second: white moving toward black:
  ```
  check(b, a, -x_dir)
  ```

If either returns true, interception is possible at step `t`, so we answer:

```
cout << "YES\n" << t << "\n";
```

and terminate.

### 2.6. Why limit to `max_steps = D/2 - 1`?

The code sets:

```
int D = max(dx, dy);
int max_steps = D / 2 - 1;
if(max_steps <= 0) {
    cout << "NO\n" << D - 1 << "\n";
    return;
}
```

Reasoning:

- White and black need about `D - 1` moves each to come within distance 1.
- By time `D/2` (roughly half that), the distance between them is already substantially reduced; in particular, if black‑white has not gotten close to the line of motion early enough, then later the squares in which they move shrink in y and narrow in x so much that there is no new intersection opportunity that wasn’t already present earlier.
- A rigorous geometric proof is nontrivial, but the intended solution observes that the earliest possible interception will necessarily be at `t ≤ D/2 - 1`. If we haven’t intercepted by then, it is too late: from then on, black & white’s paths are too constrained and close to each other for black‑white to insert itself without having had a previous opportunity.

If `max_steps <= 0`, i.e. `D ≤ 2`, then black and white start extremely close:

- They can become neighbors essentially immediately.
- Black‑white has no time for a proactive interception except possibly at t=0, but the problem disallows capture without movement.

Thus we directly answer:

```
NO
D - 1
```

since black‑white cannot win.

### 2.7. Complexity

- We perform one loop over `t` from 1 to at most `D/2 - 1`.  
  Note: D ≤ 10^6 (since coordinates in [1, N] and N ≤ 10^6).
- Each iteration constant time.  
  Overall \(O(D)\) which is \(O(N)\), acceptable for N up to 10^6 in 0.25s in C++.

Memory usage is O(1).

---

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;
};

int n;
pair<int, int> a, b, c;

void read() {
    cin >> n;
    cin >> a >> b >> c;
}

void solve() {
    // The solution to this problem is case work. Probably the hardest part is
    // to actually parse what it asks for. Essentially, we want to check if the
    // "black-white" king can intersect the path of the two other kings. The two
    // other kings will always take max(|x1-x2|, |y1-y2|) steps to meet (time is
    // divided by 2 because they make two steps for a single step of the
    // "black-white"). As N isn't large, we can iterate through the steps until
    // we meet. At step T, we can figure out the range of y coordinates that are
    // reachable by the two kings, and see if the "black-white" can be on any of
    // these cells. This solution is in O(N).

    if(a == c || b == c) {
        cout << "YES\n0\n";
        return;
    }

    int dx = abs(a.first - b.first);
    int dy = abs(a.second - b.second);
    if(dx < dy) {
        swap(a.first, a.second);
        swap(b.first, b.second);
        swap(c.first, c.second);
    }

    int D = max(dx, dy);
    int max_steps = D / 2 - 1;
    int x_dir = (b.first > a.first) ? 1 : -1;

    if(max_steps <= 0) {
        cout << "NO\n" << D - 1 << "\n";
        return;
    }

    auto clamp = [](int v, int lo, int hi) { return max(lo, min(hi, v)); };

    for(int t = 1; t <= max_steps; t++) {
        int x3 = c.first - t, x4 = c.first + t;
        int y3 = c.second - t, y4 = c.second + t;

        auto check = [&](pair<int, int> king, pair<int, int> other, int dir) {
            int x = king.first + dir * t;
            int y_min =
                max(clamp(king.second - t, 1, n),
                    clamp(other.second - D + t, 1, n));
            int y_max =
                min(clamp(king.second + t, 1, n),
                    clamp(other.second + D - t, 1, n));

            if(x < x3 || x > x4) {
                return false;
            }
            if(x == x3 || x == x4) {
                return max(y_min, y3) <= min(y_max, y4);
            }
            return (y_min <= y3 && y3 <= y_max) || (y_min <= y4 && y4 <= y_max);
        };

        if(check(a, b, x_dir) || check(b, a, -x_dir)) {
            cout << "YES\n" << t << "\n";
            return;
        }
    }

    cout << "NO\n" << D - 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**

This Python code mirrors the C++ logic as closely as possible, preserving the same geometric reasoning and time complexity.

```python
import sys
from typing import Tuple

def clamp(v: int, lo: int, hi: int) -> int:
    """Clamp integer v into [lo, hi]."""
    if v < lo:
        return lo
    if v > hi:
        return hi
    return v

def solve_one(n: int, a: Tuple[int, int], b: Tuple[int, int], c: Tuple[int, int]) -> str:
    """
    Solve one instance of the Black-white king problem.

    n: board size (N x N)
    a: (x, y) of first king (call it black)
    b: (x, y) of second king (call it white)
    c: (x, y) of black-white (invisible) king

    Returns the output string with two lines.
    """
    # If black-white king starts on same cell as one of the kings
    # (should not happen with valid input, but we keep the C++ logic)
    if a == c or b == c:
        return "YES\n0\n"

    ax, ay = a
    bx, by = b
    cx, cy = c

    # Distances in x and y between the two visible kings
    dx = abs(ax - bx)
    dy = abs(ay - by)

    # If vertical distance is larger, swap x and y for all three kings
    # to enforce dx >= dy.
    if dx < dy:
        # swap coordinates: (x, y) -> (y, x)
        ax, ay = ay, ax
        bx, by = by, bx
        cx, cy = cy, cx
        # Note: n remains same as board is square
        dx = abs(ax - bx)
        dy = abs(ay - by)

    # Chebyshev distance between the two visible kings
    D = max(dx, dy)

    # Max number of black-white steps we need to check
    max_steps = D // 2 - 1

    # Direction of motion along x for king b relative to king a
    # +1 if b is to the right of a, -1 if to the left (or same x, but then dx=0).
    x_dir = 1 if bx > ax else -1

    # If there's no positive time when interception is possible before meet
    if max_steps <= 0:
        # Minimal number of moves for each of the two visible kings
        # to become neighbors is D - 1.
        return f"NO\n{D - 1}\n"

    # Loop over time steps t = 1..max_steps
    for t in range(1, max_steps + 1):
        # Black-white king's reachable x-range and y-range
        x3 = cx - t
        x4 = cx + t
        y3 = cy - t
        y4 = cy + t

        def check(king_x: int, king_y: int, other_x: int, other_y: int, dir_: int) -> bool:
            """
            Check if a given visible king starting at (king_x, king_y),
            moving along shortest path to (other_x, other_y) with x-direction dir_,
            can be in any cell of the black-white king's reachable square
            at time t.
            """
            # x-position after t steps
            x = king_x + dir_ * t

            # y-position range after t steps, combining:
            #  1) up to t vertical steps
            #  2) constraints imposed by shortest path to (other_x, other_y)
            y_min = max(
                clamp(king_y - t, 1, n),
                clamp(other_y - D + t, 1, n)
            )
            y_max = min(
                clamp(king_y + t, 1, n),
                clamp(other_y + D - t, 1, n)
            )

            # If x is outside [x3, x4], there's no overlap in x
            if x < x3 or x > x4:
                return False

            # If x is exactly on the left or right boundary of black-white's square
            if x == x3 or x == x4:
                # Intersection if vertical ranges intersect
                return max(y_min, y3) <= min(y_max, y4)

            # If x is strictly inside the x-interval:
            # Intersection if there's some y in the overlap of [y_min,y_max] and [y3,y4].
            # Checking if y3 or y4 lies in [y_min, y_max] is sufficient.
            if y_min <= y3 <= y_max:
                return True
            if y_min <= y4 <= y_max:
                return True
            return False

        # Try intercepting king a moving toward b
        if check(ax, ay, bx, by, x_dir):
            return f"YES\n{t}\n"
        # Try intercepting king b moving toward a (opposite x direction)
        if check(bx, by, ax, ay, -x_dir):
            return f"YES\n{t}\n"

    # If for all t <= max_steps interception is impossible,
    # black-white king cannot win.
    return f"NO\n{D - 1}\n"


def main() -> None:
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)

    # Single test case as in provided C++ code
    n = int(next(it))
    a = (int(next(it)), int(next(it)))
    b = (int(next(it)), int(next(it)))
    c = (int(next(it)), int(next(it)))

    sys.stdout.write(solve_one(n, a, b, c))


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

---

5. **Compressed editorial**

- Distance between black and white kings is \(D = \max(|x_1 - x_2|, |y_1 - y_2|)\). They can become neighbors in `D-1` moves each; if black‑white cannot intercept, answer is `NO` and `D-1`.

- Black‑white moves after both visible kings each step; at black‑white’s t‑th move, each visible king has also made t moves.

- W.l.o.g. enforce \(|dx| ≥ |dy|\) by swapping x,y of all 3 if needed. Now main motion is along x.

- There exists an optimal shortest‑path meeting strategy where each of the visible kings moves monotonically along x. Let `x_dir` be direction of king `b` relative to `a` (+1 or −1).

- Black‑white king at time t can occupy any square within Chebyshev distance t of its start: `[cx − t, cx + t] × [cy − t, cy + t]`.

- For a visible king at `(kx, ky)` moving towards `(ox, oy)` in x‑direction `dir`, at time t:

  - Its x is fixed: `x = kx + dir * t`.
  - Its y is constrained by two conditions:
    1. Vertical speed: `ky - t ≤ y ≤ ky + t`.
    2. Being on some shortest path of total length D to the other king:  
       `oy - (D - t) ≤ y ≤ oy + (D - t)`.  
    After clamping to board [1, N], its feasible y‐interval is:
    ```text
    y_min = max(clamp(ky - t), clamp(oy - D + t))
    y_max = min(clamp(ky + t), clamp(oy + D - t))
    ```

- Now we need to check if this vertical segment at fixed x intersects black‑white’s t‑square. Let black‑white’s t‑square be `[x3,x4]×[y3,y4]`.

  - If `x < x3` or `x > x4`: no intersection.
  - Else if `x == x3 or x == x4`: intersection iff `[y_min,y_max]` intersects `[y3,y4]`.
  - Else (`x3 < x < x4`): intersection iff there exists y ∈ `[y_min,y_max] ∩ [y3,y4]` (e.g. if y3 or y4 lies in `[y_min,y_max]`).

- We perform this check for both visible kings for each time t.

- How long to check? After approximately `D/2` black‑white moves, the opportunity to get between the visible kings disappears. It suffices to check
  \[
  t = 1..(\lfloor D/2 \rfloor - 1).
  \]
  If this upper bound `max_steps` ≤ 0, interception is impossible; answer `NO` and `D-1`.

- Complexity is O(D) per test, D ≤ 10^6 → fast enough in C++ and in optimized Python.