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 JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.Math;


public class Main {
    static int n;
    //static double PI = Math.PI;
    static final int maxn = 100010;
    static Point a[] = new Point[maxn];
    static Point p;
    
    public static void main(String args[])  throws Exception
    {
        init();
        doit();
    }
    
    public static double calPointToLine(Point a, Point b, Point p)//线段ab到点p距离
    {
        Line A = new Line(a, b);
        Point jiaoDian = CalSomeProblem.nearesPointToLineSeg(p, A);
        //System.out.println(jiaoDian.x + " " + jiaoDian.y);
        return Point.dist(jiaoDian, p);
    }
    
    public static void doit()
    {
        //for (int i = 0; i != n ; ++ i)    System.out.println(a[i].x +" " + a[i].y);
        
        double max = -1000000000000000000l;
        double min =calPointToLine(a[0],a [n-1], p); 
        
        for (int i = 0; i != n; ++ i)
        {
            double tmp = Point.dist(a[i], p);
            if (tmp > max)  max = tmp;
            if (i != n -1)
            {
                double tmp2 = calPointToLine(a[i], a[i + 1], p);
                if (tmp2 < min) min = tmp2;
            }
        }
        //System.out.println("maxn si "+ max);
        //System.out.println("min si "+ min);
        System.out.printf("%.10f",Math.PI * (max * max - min * min) / 10000);
    }
    
    public static void init()
    {
        Scanner cin = new Scanner(System.in);
        n = cin.nextInt();
        p = new Point(cin.nextLong() * 100, cin.nextLong() * 100);
        for (int i = 0; i != n; ++ i)
        {
            a[i] = new Point(cin.nextDouble() * 100 ,cin.nextDouble() * 100);
        }
        cin.close();
    }
}

class Point
{
    public double x, y;
    public Point()
    {
        x = 0;
        y = 0;
    }
    public Point (double a, double b)
    {
        x = a;
        y = b;
    }
    
    //坐标相减
    Point dec(Point b)
    {
        return new Point(x-b.x, y-b.y);
    }
    public static Point dec(Point a, Point b)
    {
        return new Point(a.x - b.x, a.y - b.y);
    }

    //叉积
    double chaJi(Point b)
    {
            return x * b.y - y * b.x;
    }
    public static double chaJi(Point a, Point b)
    {
        return a.x * b.y - a.y * b.x;
    }

    //点积
    double dianJi(Point b)
    {
        return x *b.x + y * b.y;
    }
    public static double dianJi(Point a, Point b)
    {
        return a.x * b.x + a.y * b.y;
    }
    
    //绕远点旋转角度B（弧度），后x,y的值
    void transXY(double B)
    {
        double tx = x, ty = y;
        x = tx * Math.cos(B) - ty * Math.sin(B);
        y = tx * Math.sin(B) + ty * Math.cos(B);
    }
    public static Point transXY(Point a, double B)
    {
        double tx = a.x, ty = a.y;
        a.x = tx * Math.cos(B) - ty * Math.sin(B);
        a.y = tx * Math.sin(B) + ty * Math.cos(B);
        return a;
    }
    
    public static double dist(Point a, Point b)
    {
        return Math.sqrt(dianJi(dec(a,b),dec(a,b)));
    }
}

class JiaodianAns //和Line联用
{
    int flag;
    Point jiaoDian;
    public JiaodianAns(){}
    public JiaodianAns(int a, Point b)
    {
        flag = a;
        jiaoDian = b;
    }
    //两直线相交求交点
    //第一个值为0标示直线垂直，为1标示平行，为0标示相交，为2标示相交
    //只有第一个值为2时，交点才有意义
}

class Line
{
    Point s, e;
    Line(){}

    Line(Point a, Point b)
    {
        s = a;
        e = b;
    }
    int sgn(double x)
    {
        if (Math.abs(x) < 1e-8) return 0;
        if (x < 0)  return -1;
        return 1;
    }
    //两直线相交求交点
    //第一个值为0标示直线垂直，为1标示平行，为0标示相交，为2标示相交
    //只有第一个值为2时，交点才有意义
    JiaodianAns isXiangJiao(Line b)
    {
        Point res = s;
        if (sgn(Point.chaJi(Point.dec(s, e),Point.dec(b.s, b.e)))==0)
        {
            if (0 == sgn(Point.chaJi(Point.dec(s, b.e), Point.dec(b.s, b.e))))
                return new JiaodianAns(0, res);
            else return new JiaodianAns(1, res);
        }
        double tmp1 = Point.chaJi(Point.dec(s, b.s), Point.dec(b.s, b.e));
        double tmp2 = Point.chaJi(Point.dec(s, e), Point.dec(b.s, b.e));
        double t = tmp1 / tmp2;
        res.x += (e.x - s.x) * t;
        res.y += (e.y - s.y) * t;
        return new JiaodianAns(2, res);
    }
}

class CalSomeProblem
{
    //计算点到【线段】的最短距离，返回最近的点
    public static Point nearesPointToLineSeg(Point P, Line L)
    {
        Point result = new Point();
        double tmp1 = Point.dianJi(Point.dec(P, L.s), Point.dec(L.e, L.s));
        double tmp2 = Point.dianJi(Point.dec(L.e, L.s), Point.dec(L.e, L.s));
        double t = tmp1 / tmp2;
        if (t >= 0 && t <= 1)
        {
            result.x = L.s.x + (L.e.x - L.s.x) * t;
            result.y = L.s.y + (L.e.y - L.s.y) * t;
        }
        else
        {
            if (Point.dist(P, L.s) < Point.dist(P, L.e))
                result = L.s;
            else result = L.e;
        }
        return result;
    }
}