CodeForces 614C. Peter and Snow Blower

C. Peter and Snow Blower

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 (), 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 .

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:

题意：求一个多边形绕多边形外的一个点转能形成的面积。 思路：可以看出形成的面积是一个环状。所以要求多边形与点p最短的距离和最远的距离，求这两个距离形成的圆形的面积的差。 最远的距离：一定是多边形上的某点到点p的距离 最短的距离：可能是多边形上的点或边到点p的距离 有公式就方便多了： 点到线段的最短距离

#include 
#define maxn 100010
using namespace std;
const double pai = 3.14159265;
struct node
{
    double x, y;
}ver[maxn];
double dist(double x1, double y1, double x2, double y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double PointToSegDist(double x, double y, double x1, double y1, double x2, double y2)
{
    double cross = (x2 - x1) * (x - x1) + (y2 - y1) * (y - y1);
    if (cross <= 0) return sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));

    double d2 = (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
    if (cross >= d2) return sqrt((x - x2) * (x - x2) + (y - y2) * (y - y2));

    double r = cross / d2;
    double px = x1 + (x2 - x1) * r;
    double py = y1 + (y2 - y1) * r;
    return sqrt((x - px) * (x - px) + (y - py) * (y - py));
}
int main()
{
    double x, y, maxlen, minlen, area;
    int n, i;
    scanf("%d %lf %lf", &n, &x, &y);
    maxlen = 0;
    minlen = 999999999;
    for(i = 0;i < n;i++){
        scanf("%lf %lf", &ver[i].x, &ver[i].y);
        maxlen = max(maxlen, dist(x, y, ver[i].x, ver[i].y));
    }

    minlen = min(minlen, PointToSegDist(x, y, ver[0].x, ver[0].y, ver[n-1].x, ver[n-1].y));
    for(i = 1;i < n;i++) minlen = min(minlen, PointToSegDist(x, y, ver[i].x, ver[i].y, ver[i-1].x, ver[i-1].y));

    area = maxlen*maxlen*pai - minlen*minlen*pai;
    printf("%lf\n", area);
}