Problem: Problem

Find the area of ​​a regular N / K polygon inscribed in a circle with a radius of 1.

However, a regular N / K polygon is defined as "the outermost figure that takes N points on the circumference at equal intervals and connects each point every K-1".

For example, a 5/2 polygon can be drawn as follows. First, take five points at equal intervals on the circumference of radius 1.

Sample Input 2 diagram


Next, connect each point every other 2-1 = 1.

Sample Input 2 diagram


The outermost figure is a regular 5/2 square.

Sample Input 2 diagram

Constraints

The input satisfies the following constraints.

* 5 ≤ N ≤ 106
* 1 <K <N / 2
* N and K are integers that are relatively prime

Input

The input is given in the following format.


N K


Two integers N and K are given on one line.

Output

Output the area of ​​a regular N / K polygon inscribed in a circle with a radius of 1 on one line. An error of 10-5 or less is acceptable.

Examples

Input

5 2


Output

1.12256994


Input

20 3


Output

2.93114293


Input

7 3


Output

1.08395920


Input

100000 3


Output

3.14159265

Solution (in CPP):

Solution:
#include "bits/stdc++.h"

using namespace std;

using ld = long double;
using Point = std::complex<ld>;

const ld eps = 1e-9, pi = acos(-1.0);

namespace std
{
bool operator<(const Point &lhs, const Point &rhs)
{
    if (lhs.real() < rhs.real() - eps)
        return true;
    if (lhs.real() > rhs.real() + eps)
        return false;
    return lhs.imag() < rhs.imag();
}
} // namespace std

Point input_point()
{
    ld x, y;
    std::cin >> x >> y;
    return Point(x, y);
}

bool eq(ld a, ld b)
{
    return (abs(a - b) < eps);
}

ld dot(Point a, Point b)
{
    return real(conj(a) * b);
}

ld cross(Point a, Point b)
{
    return imag(conj(a) * b);
}

// CCW::counter clockwise
int ccw(Point a, Point b, Point c)
{
    b -= a;
    c -= a;
    if (cross(b, c) > eps)
        return 1; // a,b,c : counter-clockwise
    if (cross(b, c) < -eps)
        return -1; // a,b,c : clockwise
    if (dot(b, c) < 0)
        return 2; // c,a,b : on a line
    if (norm(b) < norm(c))
        return -2; // a,b,c : on a line
    return 0;      // a,c,b : on a line
}

class Line
{
public:
    Point a, b;
    Line() : a(Point(0, 0)), b(Point(0, 0)) {}
    Line(Point a, Point b) : a(a), b(b) {}
};

ld dot(Line l, Line m)
{
    return dot((l.a - l.b), (m.a - m.b));
}

// l:line, m:line が交点を持つか
bool isis_ll(Line l, Line m)
{
    return !eq(cross(l.b - l.a, m.b - m.a), 0);
}

// l:line, s:segment
bool isis_ls(Line l, Line s)
{
    return isis_ll(l, s) &&
           (cross(l.b - l.a, s.a - l.a) * cross(l.b - l.a, s.b - l.a) < eps);
}

// s:segment, t:segment
bool isis_ss(Line s, Line t)
{
    return ccw(s.a, s.b, t.a) * ccw(s.a, s.b, t.b) <= 0 &&
           ccw(t.a, t.b, s.a) * ccw(t.a, t.b, s.b) <= 0;
}

// p が l:line 上に存在するか
bool isis_lp(Line l, Point p)
{
    return (abs(cross(l.b - p, l.a - p)) < eps);
}

bool isis_sp(Line s, Point p)
{
    return (abs(s.a - p) + abs(s.b - p) - abs(s.b - s.a) < eps);
}

// p から l に下ろした足との交点
Point proj(Line l, Point p)
{
    ld t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
    return l.a + t * (l.a - l.b);
}

// l:line, t:line の交点
Point is_ll(Line l, Line m)
{
    Point lv = l.b - l.a, mv = m.b - m.a;
    assert(cross(lv, mv) != 0);
    return l.a + lv * cross(mv, m.a - l.a) / cross(mv, lv);
}

// p, l:line の距離
ld dist_lp(Line l, Point p)
{
    return abs(p - proj(l, p));
}

ld dist_ll(Line l, Line m)
{
    return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}

ld dist_ls(Line l, Line s)
{
    return isis_ls(l, s) ? 0 : std::min(dist_lp(l, s.a), dist_lp(l, s.b));
}

ld dist_sp(Line s, Point p)
{
    Point r = proj(s, p);
    return isis_sp(s, r) ? abs(r - p) : std::min(abs(s.a - p), abs(s.b - p));
}

ld dist_ss(Line s, Line t)
{
    if (isis_ss(s, t))
        return 0;
    return std::min({dist_sp(s, t.a), dist_sp(s, t.b), dist_sp(t, s.a), dist_sp(t, s.b)});
}

// a, b の垂直二等分線. a -> b を90度反時計回り回転
Line bisector(Point a, Point b)
{
    Point mid = (a + b) * Point(0.5, 0);
    return Line(mid, mid + (b - a) * Point(0, pi / 2));
}

// 直線 l, m のなす角を求める
ld degree_ll(Line l, Line m)
{
    ld cos_shita = dot(l, m) / (abs(l.b - l.a) * abs(m.b - m.a));
    if (cos_shita < -1.0)
        cos_shita = -1.0;
    if (cos_shita > 1.0)
        cos_shita = 1.0;
    ld shita = acos(cos_shita);
    // shita = sita * 180.0 / PI;
    return shita;
}

using Polygon = std::vector<Point>;

ld area(const Polygon &p)
{
    ld res = 0;
    int n = p.size();
    for (int i = 0; i < n; i++)
    {
        res += cross(p[i], p[(i + 1) % n]);
    }
    return res / 2;
}

template <typename T>
void printv(const vector<T> &v)
{
    int sz = v.size();
    for (int i = 0; i < sz; i++)
    {
        cout << v[i] << " \n"[i == sz - 1];
    }
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;
    vector<Point> ps;
    for (int i = 0; i < n; i++)
    {
        ld x = cos(2 * pi * i / n), y = sin(2 * pi * i / n);
        ps.push_back(Point(x, y));
    }
    Line l0 = Line(ps[0], ps[k]);
    Line l1 = Line(ps[1], ps[n - k + 1]);
    Point p = is_ll(l0, l1);
    //cout << p.real() << " " << p.imag() << endl;
    Polygon poly = {Point(0, 0), ps[0], p};
    cout << fixed << setprecision(10) << area(poly) * 2 * n << endl;
}