Problem: Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.

Formally, we assume that Peter's machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.

Peter decided to tie his car to point P and now he is wondering what is the area of ​​the region that will be cleared from snow. Help him.

Input

The first line of the input contains three integers — the number of vertices of the polygon n (<image>), and coordinates of point P.

Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.

All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.

Output

Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6. 

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.

Examples

Input

3 0 0
0 1
-1 2
1 2


Output

12.566370614359172464


Input

4 1 -1
0 0
1 2
2 0
1 1


Output

21.991148575128551812

Note

In the first sample snow will be removed from that area:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = 1e20;
const int maxp = 1111;
int dblcmp(double d) {
  if (fabs(d) < eps) return 0;
  return d > eps ? 1 : -1;
}
inline double sqr(double x) { return x * x; }
struct point {
  double x, y;
  point() {}
  point(double _x, double _y) {
    x = _x;
    y = _y;
  }
  void input() { scanf("%lf%lf", &x, &y); }
  void output() { printf("%.2f %.2f\n", x, y); }
  bool operator==(point a) const {
    return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;
  }
  bool operator<(point a) const {
    return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x;
  }
  point operator-(point a) const { return point(x - a.x, y - a.y); }
  double len() { return hypot(x, y); }
  double len2() { return x * x + y * y; }
  double distance(point p) { return hypot(x - p.x, y - p.y); }
  point add(point p) { return point(x + p.x, y + p.y); }
  point sub(point p) { return point(x - p.x, y - p.y); }
  point mul(double b) { return point(x * b, y * b); }
  point div(double b) { return point(x / b, y / b); }
  double dot(point p) { return x * p.x + y * p.y; }
  double det(point p) { return x * p.y - y * p.x; }
  double rad(point a, point b) {
    point p = *this;
    return fabs(atan2(fabs(a.sub(p).det(b.sub(p))), a.sub(p).dot(b.sub(p))));
  }
  point trunc(double r) {
    double l = len();
    if (!dblcmp(l)) return *this;
    r /= l;
    return point(x * r, y * r);
  }
  point rotleft() { return point(-y, x); }
  point rotright() { return point(y, -x); }
  point rotate(point p, double angle) {
    point v = this->sub(p);
    double c = cos(angle), s = sin(angle);
    return point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
  }
};
struct line {
  point a, b;
  line() {}
  line(point _a, point _b) {
    a = _a;
    b = _b;
  }
  bool operator==(line v) { return (a == v.a) && (b == v.b); }
  line(point p, double angle) {
    a = p;
    if (dblcmp(angle - pi / 2) == 0) {
      b = a.add(point(0, 1));
    } else {
      b = a.add(point(1, tan(angle)));
    }
  }
  line(double _a, double _b, double _c) {
    if (dblcmp(_a) == 0) {
      a = point(0, -_c / _b);
      b = point(1, -_c / _b);
    } else if (dblcmp(_b) == 0) {
      a = point(-_c / _a, 0);
      b = point(-_c / _a, 1);
    } else {
      a = point(0, -_c / _b);
      b = point(1, (-_c - _a) / _b);
    }
  }
  void input() {
    a.input();
    b.input();
  }
  void adjust() {
    if (b < a) swap(a, b);
  }
  double length() { return a.distance(b); }
  double angle() {
    double k = atan2(b.y - a.y, b.x - a.x);
    if (dblcmp(k) < 0) k += pi;
    if (dblcmp(k - pi) == 0) k -= pi;
    return k;
  }
  int relation(point p) {
    int c = dblcmp(p.sub(a).det(b.sub(a)));
    if (c < 0) return 1;
    if (c > 0) return 2;
    return 3;
  }
  bool pointonseg(point p) {
    return dblcmp(p.sub(a).det(b.sub(a))) == 0 &&
           dblcmp(p.sub(a).dot(p.sub(b))) <= 0;
  }
  bool parallel(line v) { return dblcmp(b.sub(a).det(v.b.sub(v.a))) == 0; }
  int segcrossseg(line v) {
    int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
    int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
    int d3 = dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
    int d4 = dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
    if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;
    return (d1 == 0 && dblcmp(v.a.sub(a).dot(v.a.sub(b))) <= 0 ||
            d2 == 0 && dblcmp(v.b.sub(a).dot(v.b.sub(b))) <= 0 ||
            d3 == 0 && dblcmp(a.sub(v.a).dot(a.sub(v.b))) <= 0 ||
            d4 == 0 && dblcmp(b.sub(v.a).dot(b.sub(v.b))) <= 0);
  }
  int segcrossseg_inside(line v) {
    if (v.pointonseg(a) || v.pointonseg(b) || pointonseg(v.a) ||
        pointonseg(v.b))
      return 0;
    int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
    int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
    int d3 = dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
    int d4 = dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
    if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 1;
    return (d1 == 0 && dblcmp(v.a.sub(a).dot(v.a.sub(b))) <= 0 ||
            d2 == 0 && dblcmp(v.b.sub(a).dot(v.b.sub(b))) <= 0 ||
            d3 == 0 && dblcmp(a.sub(v.a).dot(a.sub(v.b))) <= 0 ||
            d4 == 0 && dblcmp(b.sub(v.a).dot(b.sub(v.b))) <= 0);
  }
  int linecrossseg(line v) {
    int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
    int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
    if ((d1 ^ d2) == -2) return 2;
    return (d1 == 0 || d2 == 0);
  }
  int linecrossline(line v) {
    if ((*this).parallel(v)) {
      return v.relation(a) == 3;
    }
    return 2;
  }
  point crosspoint(line v) {
    double a1 = v.b.sub(v.a).det(a.sub(v.a));
    double a2 = v.b.sub(v.a).det(b.sub(v.a));
    return point((a.x * a2 - b.x * a1) / (a2 - a1),
                 (a.y * a2 - b.y * a1) / (a2 - a1));
  }
  double dispointtoline(point p) {
    return fabs(p.sub(a).det(b.sub(a))) / length();
  }
  double dispointtoseg(point p) {
    if (dblcmp(p.sub(b).dot(a.sub(b))) < 0 ||
        dblcmp(p.sub(a).dot(b.sub(a))) < 0) {
      return min(p.distance(a), p.distance(b));
    }
    return dispointtoline(p);
  }
  point lineprog(point p) {
    return a.add(b.sub(a).mul(b.sub(a).dot(p.sub(a)) / b.sub(a).len2()));
  }
  point symmetrypoint(point p) {
    point q = lineprog(p);
    return point(2 * q.x - p.x, 2 * q.y - p.y);
  }
};
pair<long long, long long> pp[100001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, x, y;
  cin >> n >> x >> y;
  double mn = 1e20, mx = 0.0;
  for (int i = (0); i < (n); ++i) {
    long long x1, y1;
    cin >> x1 >> y1;
    x1 -= x;
    y1 -= y;
    pp[i] = make_pair(x1, y1);
  }
  for (int i = (0); i < (n); ++i) {
    int j = (i + 1) % n;
    line l1 = line(point(pp[i].first, pp[i].second),
                   point(pp[j].first, pp[j].second));
    double v = l1.dispointtoseg(point(0, 0));
    mx = max(mx, v);
    mn = min(mn, v);
    v = point(0, 0).distance(l1.a);
    mx = max(mx, v);
    mn = min(mn, v);
  }
  double v = mx * mx - mn * mn;
  if (n == 1) {
    v = 2.0 * mx;
  }
  cout << fixed << setprecision(10) << v * acos(-1.0) << endl;
}