Problem: The first ship with the Earth settlers landed on Mars. The colonists managed to build n necessary structures on the surface of the planet (which can be regarded as a plane, and the construction can be regarded as points on it). But one day the scanners recorded suspicious activity on the outskirts of the colony. It was decided to use the protective force field generating system to protect the colony against possible trouble.

The system works as follows: the surface contains a number of generators of the field (they can also be considered as points). The active range of each generator is a circle of radius r centered at the location of the generator (the boundary of the circle is also included in the range). After the system is activated, it stretches the protective force field only over the part of the surface, which is within the area of all generators' activity. That is, the protected part is the intersection of the generators' active ranges.

The number of generators available to the colonists is not limited, but the system of field generation consumes a lot of energy. More precisely, the energy consumption does not depend on the number of generators, but it is directly proportional to the area, which is protected by the field. Also, it is necessary that all the existing buildings are located within the protected area.

Determine the smallest possible area of the protected part of the surface containing all the buildings.

Input

The first line contains two integers n and r (1 ≤ n ≤ 105, 1 ≤ r ≤ 50000) — the number of buildings and the active ranges of the generators, correspondingly.

Next n lines contains the buildings' coordinates. The i + 1-th (1 ≤ i ≤ n) line contains two real numbers with at most three digits after the decimal point xi and yi (|xi|, |yi| ≤ 50000) — coordinates of the i-th building. It is guaranteed that no two buildings are located at the same point, and no two different buildings are located closer than 1.

It is guaranteed that there exists a circle with radius r that contains all the buildings.

Output

Print the single real number — the minimum area of the protected part containing all the buildings. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.

Examples

Input

3 5
0.00 0.000
0.0 8.00
6 8.00


Output

78.5398163397


Input

4 1000
0.0 0.0
0 2.00
2.00 2
2.0 0.00


Output

4.0026666140


Input

4 5
3.00 0.0
-3 0.00
0.000 1
0.0 -1.00


Output

8.1750554397

Note

In the first sample the given radius equals the radius of the circle circumscribed around the given points. That's why the circle that corresponds to it is the sought area. The answer is 25π.

In the second sample the area nearly coincides with the square which has vertexes in the given points.

The area for the third sample is shown on the picture below.

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, const U &b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, const U &b) {
  if (a < b) a = b;
}
struct point {
  double first, second;
  point(double first = 0, double second = 0) : first(first), second(second) {}
  inline double operator^(point a) {
    return first * a.second - second * a.first;
  }
  inline double operator*(point a) {
    return first * a.first + second * a.second;
  }
  inline point operator*(double a) { return point(first * a, second * a); }
  inline point operator/(double a) { return point(first / a, second / a); }
  inline point operator-(point a) {
    return point(first - a.first, second - a.second);
  }
  inline point operator+(point a) {
    return point(first + a.first, second + a.second);
  }
  inline point rotate(double al) { return *this & point(cos(al), sin(al)); }
  inline point operator&(point a) {
    return point(first * a.first - second * a.second,
                 first * a.second + second * a.first);
  }
  inline point norm() {
    double s = nor();
    if (s < 1e-7) return *this;
    return point(first / s, second / s);
  }
  inline int pos() {
    return second > 1e-7 || fabs(second) < 1e-7 && first > -1e-7;
  }
  inline double nor() { return sqrt(nor2()); }
  inline double nor2() { return *this * *this; }
  inline point conj() { return point(-second, first); }
  inline double al() { return atan2(second, first); }
  inline bool operator<(const point &a) const {
    if (fabs(first - a.first) > 1e-7) return first < a.first;
    return second < a.second - 1e-7;
  }
  inline void input() { scanf("%lf %lf", &first, &second); }
  inline void print() { printf("%lf %lf\n", first, second); }
};
inline int sgn(double a) { return (a > 1e-7) - (a < -1e-7); }
inline bool is_on(point a, point b, point c) {
  if (sgn(a - b ^ b - c)) return 0;
  return sgn((a - c) * (b - c)) <= 0;
}
inline double angle(point a, point b) {
  double res = a * b / a.nor() / b.nor();
  res = acos(res);
  return res;
}
inline double angle(point a, point b, point c) { return angle(a - b, c - b); }
int n;
double r, ans;
point p[100000];
inline bool in(point a, point b, point c) {
  point M = (b + c) / 2;
  point dir = (c - b).conj().norm();
  point O = M + dir * sqrt(r * r - (b - M).nor2());
  return (O - a).nor() < r + 1e-7;
}
inline double fan_size(point a, point b) {
  point M = (a + b) / 2,
        dir = (b - a).conj().norm() * sqrt(r * r - (b - M).nor2());
  point O = M + dir;
  double ang = angle(a, O, b);
  return (r * r * ang - (a - b).nor() * (M - O).nor()) / 2;
}
int main() {
  scanf("%d %lf", &n, &r);
  if (n == 1) return puts("0");
  for (int i = 0; i < n; i++) {
    p[i].input();
    if (p[i] < p[0]) swap(p[0], p[i]);
  }
  sort(p + 1, p + n, [](point a, point b) {
    a = a - p[0], b = b - p[0];
    int sg = sgn(a ^ b);
    if (sg) return sg > 0;
    return a.nor() < b.nor();
  });
  int N = 0;
  for (int i = 0; i < n; i++) {
    while (N > 1 && (p[N - 1] - p[i] ^ p[N - 2] - p[i]) > -1e-7) N--;
    p[N++] = p[i];
  }
  n = 0;
  for (int i = 0; i < N; i++) {
    while (n > 1 && in(p[n - 1], p[n - 2], p[i])) n--;
    p[n++] = p[i];
  }
  p[n] = p[0];
  int st = 0, ed = n - 1;
  while (st + 1 < ed) {
    if (in(p[ed], p[ed - 1], p[st]))
      ed--;
    else if (in(p[st], p[ed], p[st + 1]))
      st++;
    else
      break;
  }
  p[ed + 1] = p[st];
  for (int i = st; i <= ed; i++) ans += p[i] ^ p[i + 1];
  ans /= 2;
  for (int i = st; i <= ed; i++) ans += fan_size(p[i], p[i + 1]);
  printf("%.5lf", ans);
}