1. Abridged Problem Statement
Given a regular N-gon with vertices numbered 1…N in clockwise order, you know the 2D coordinates of two distinct vertices indices N1 and N2. Compute and print the coordinates of all N vertices in index order, to 6 decimal places.

2. Detailed Editorial
We have a regular N-gon inscribed in a circle. Let its center be C and radius R. Two known vertices A and B correspond to indices a and b (1≤a,b≤N). The central angle between A and B along the circumcircle is
  Δ = 2π·|b−a|/N.
The straight‐line distance d = |A−B| is the chord length for that central angle. By the chord–radius relation,
  d = 2R·sin(Δ/2) ⇒ R = d / (2·sin(Δ/2)).

To find the center C, observe that C lies on the perpendicular bisector of segment AB, at distance h from the midpoint M of AB, where
  h = R·cos(Δ/2) = (d/2)/tan(Δ/2) .
Compute M = (A+B)/2 and a perpendicular direction to vector AB, e.g. rotate AB by +90°: (dx,dy)→(dy,−dx). Scaling that perpendicular (whose length equals d) by h/d and adding it to M gives the center with the orientation consistent with clockwise numbering.

Once C and R are known, compute the polar angle φ of point A relative to C. The reference solution recovers φ from asin((A.y−C.y)/R) and corrects the quadrant using acos((A.x−C.x)/R); this is equivalent to atan2(A.y−C.y, A.x−C.x). Then, because vertices are numbered clockwise, the i-th vertex has angle
  φ_i = φ + 2π·(a−i)/N.

Finally,
  P_i = C + R·(cos φ_i, sin φ_i).

All computations take O(N), trivially small for N≤150.

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;
};

using Point = complex<long double>;

int n, a, b;
vector<Point> points;

void read() {
    cin >> n >> a >> b;
    points.assign(n + 1, Point());
    long double x, y;
    cin >> x >> y;
    points[a] = Point(x, y);
    cin >> x >> y;
    points[b] = Point(x, y);
}

void solve() {
    // The islands are vertices of a regular N-gon listed clockwise, and we know
    // two of them (indices a, b). The chord between vertices a and b subtends a
    // central angle of 2*pi*(b-a)/n, so the circumradius is dist / 2 /
    // sin(pi*(b-a)/n) and the circle centre lies on the perpendicular bisector
    // of that chord (offset by the half-chord over tan of the half-angle).
    // Once the centre and the angular position phi of vertex a are known, every
    // other vertex i sits at angle phi + 2*pi*(a-i)/n on the circle.

    if(a > b) {
        swap(a, b);
    }

    const long double PI = 3.14159265358979323846L;
    long double dist = abs(points[b] - points[a]);
    long double radius = dist / sin(PI * (b - a) / n) / 2;

    Point mid = (points[a] + points[b]) / 2.0L;
    Point center =
        mid +
        Point(
            (points[b].imag() - points[a].imag()) / tan(PI * (b - a) / n) / 2,
            -(points[b].real() - points[a].real()) / tan(PI * (b - a) / n) / 2
        );

    long double phi = asin((points[a].imag() - center.imag()) / radius);
    if(acos((points[a].real() - center.real()) / radius) > PI / 2) {
        phi = (phi >= 0 ? PI - phi : -PI - phi);
    }

    for(int i = 1; i <= n; i++) {
        if(i != a && i != b) {
            long double delta = phi + 2 * PI * (a - i) / n;
            points[i] =
                center + Point(radius * cos(delta), radius * sin(delta));
        }
    }

    cout << fixed << setprecision(6);
    for(int i = 1; i <= n; i++) {
        cout << points[i].real() << " " << points[i].imag() << "\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 math
import sys

def main():
    data = sys.stdin.read().strip().split()
    n, a, b = map(int, data[:3])
    xa, ya = map(float, data[3:5])
    xb, yb = map(float, data[5:7])

    # Ensure a < b by swapping if needed
    if a > b:
        a, b = b, a
        xa, xb = xb, xa
        ya, yb = yb, ya

    # Convert to complex for convenience
    A = complex(xa, ya)
    B = complex(xb, yb)

    # Central half-angle between A and B
    PI = math.pi
    half_angle = PI * (b - a) / n

    # Chord length
    d = abs(B - A)

    # Radius from chord relation d = 2R sin(half_angle)
    R = d / (2 * math.sin(half_angle))

    # Midpoint of AB
    M = (A + B) / 2

    # Distance from midpoint to center along perpendicular bisector
    h = (d / 2) / math.tan(half_angle)

    # Perpendicular direction: rotate AB by +90°, i.e. (dx,dy)->(dy,-dx)
    dx = B.real - A.real
    dy = B.imag - A.imag
    # Normalize the perpendicular vector length to h
    perp = complex(dy * (h / d), -dx * (h / d))

    # Choose the correct center (the one that yields clockwise numbering).
    # We take M + perp, which matches the C++ orientation.
    C = M + perp

    # Base angle φ for vertex a
    phi = math.atan2(A.imag - C.imag, A.real - C.real)

    # Precompute step for each index shift (clockwise)
    step = 2 * PI / n

    # Generate and print all vertices
    out = []
    for i in range(1, n + 1):
        if i == a:
            P = A
        elif i == b:
            P = B
        else:
            # For clockwise numbering, angle = φ + step*(a - i)
            ang = phi + step * (a - i)
            P = C + R * complex(math.cos(ang), math.sin(ang))
        out.append(f"{P.real:.6f} {P.imag:.6f}")

    sys.stdout.write("\n".join(out))

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

5. Compressed Editorial
Compute the circumcircle of the regular N-gon using the given chord AB:
1. Let Δ = 2π·|b−a|/N, chord length d = |A−B|.
2. Radius R = d/(2·sin(Δ/2)).
3. Center C lies on the perpendicular bisector of AB at distance h=(d/2)/tan(Δ/2) from the midpoint.
4. Find φ, the polar angle of A about C.
5. Generate each vertex i by φ_i=φ+2π·(a−i)/N and P_i = C + R·(cosφ_i, sinφ_i).
