You are a programmer working for your local city hall. The town you live in has a thriving tourism industry, especially the beaches on the shores of remote islands. The area around this island is composed of sandy beaches, but in recent years the area of ​​the sandy beaches has decreased due to coastal erosion. The Tourism Division, which took the situation seriously, asked you to simulate the shape change of the beach.

In the simulation, the shape of the beach after a certain period of time is obtained based on the current shape of the beach. For simplicity, the shape of the island's coastline is considered polygonal (not necessarily convex). During the period, it is assumed that the beach is eroded by R at a distance from the current coastline in Manhattan and sinks into the sea. Here, the Manhattan distance is the sum of the difference between the x-coordinates and the difference between the y-coordinates. That is, the Manhattan distance between (x1, y1) and (x2, y2) is given by | x1 --x2 | + | y1 --y2 |.

Your job is to write a program to find the length of the shoreline obtained as a result of the simulation from the shoreline of the sandy beach given as input. If an island is divided into two or more due to coastal erosion, find the length of the coastline for each divided island and output the sum of them.

It should be noted that the two line segments included in the coastline are parallel and the Manhattan distance is not exactly 2R.

Input

The input consists of multiple datasets. Each dataset is given in the following format.

> N R
> x1 y1
> x2 y2
> ...
> xN yN

The first line gives two non-negative integers N (3 ≤ N ≤ 100) and R (0 <R ≤ 100). These integers represent the number of vertices on the coastline and the distance eroded, respectively. In the following N lines, vertex information is given to each line, one for each line. Vertex information is given by two integers xi, yi (-10000 ≤ xi, yi ≤ 10000). The coordinates (xi, yi) represent the coordinates of the i-th vertex of the coastline, respectively. The coordinates of the vertices of the island are always given counterclockwise.

The end of the input is indicated by a single line containing two zeros separated by blanks.

Output

Output the length of the eroded coastline for each dataset. If the island is completely eroded and disappears, the length of the coastline should be treated as 0.0. The output value may contain an error of 0.01 or less. In addition, the value may be displayed in any number of digits after the decimal point.

Sample Input


3 1
0 0
10 0
5 10
3 1
0 0
10 0
0 10
4 1
0 0
10 0
10 10
0 10
0 0


Output for the Sample Input


22.6524758425
23.8994949366
32.0000000000






Example

Input

3 1
0 0
10 0
5 10
3 1
0 0
10 0
0 10
4 1
0 0
10 0
10 10
0 10
0 0


Output

22.6524758425
23.8994949366
32.0000000000
#include<bits/stdc++.h>
#define inf 1<<29
#define linf 1e16
#define eps (1e-5)
#define mod 1000000007
#define pi acos(-1)
#define phi (1.0+sqrt(5))/2.0
#define f first
#define s second
#define mp make_pair
#define pb push_back
#define all(a) (a).begin(),(a).end()
#define pd(a) printf("%.10f\n",(double)(a))
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define RFOR(i,a,b) for(int i=(a)-1;(b)<=i;i--)
#define equals(a,b) (fabs((a)-(b))<eps)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<int,double> pid;
typedef pair<double,int> pdi;
typedef vector<int> vi;
typedef vector<pii> vpi;
const int dx[8]={1,0,-1,0,1,1,-1,-1};
const int dy[8]={0,1,0,-1,1,-1,1,-1};

class Point{
public:
  double x,y;
  Point(double x=0,double y=0):x(x),y(y){}

  Point operator+(Point p){ return Point(x+p.x,y+p.y);}
  Point operator-(Point p){ return Point(x-p.x,y-p.y);}
  Point operator*(double k){ return Point(x*k,y*k);}
  Point operator/(double k){ return Point(x/k,y/k);}
  bool operator<(Point p)const{ 
       return x!=p.x ? x<p.x : y<p.y; }
  bool operator==(Point p)const{ 
       return fabs(x-p.x)<eps && fabs(y-p.y)<eps;}

  double abs(){ return sqrt(norm());}
  double norm(){ return (x*x+y*y);}
};
typedef Point Vector;
typedef vector<Point> Polygon;

class Segment{
public:
  Point p1,p2;
  Segment(Point p1=Point(),Point p2=Point()):p1(p1),p2(p2){}
};
typedef Segment Line;

double norm(Vector a){ return (a.x*a.x+a.y*a.y);}
double abs(Vector a){ return sqrt(norm(a));}
double dot(Vector a,Vector b){ return (a.x*b.x+a.y*b.y);}
double cross(Vector a,Vector b){ return (a.x*b.y-a.y*b.x);}

bool isParallel(Vector a,Vector b){
    return equals(cross(a,b),0.0);
}

bool isParallel(Segment s,Segment t){
    return equals(cross(s.p1-s.p2,t.p1-t.p2),0.0);
}

int ccw(Point p0,Point p1,Point p2){
  Vector a=p1-p0;
  Vector b=p2-p0;
  if(cross(a,b)>eps)return 1;
  if(cross(a,b)<-eps)return -1;
  if(dot(a,b)<-eps)return 2;
  if(a.norm()<b.norm())return -2;
  return 0;
}

bool intersect(Point p1,Point p2,Point p3,Point p4){
  return (ccw(p1,p2,p3)*ccw(p1,p2,p4)<=0 &&
          ccw(p3,p4,p1)*ccw(p3,p4,p2)<=0);
}

bool intersect(Segment s1,Segment s2){
  return intersect(s1.p1,s1.p2,s2.p1,s2.p2);
}

Point getCrossPointLL(Line a,Line b){
  double A=cross(a.p2-a.p1,b.p2-b.p1);
  double B=cross(a.p2-a.p1,a.p2-b.p1);
  if(abs(A)<eps || abs(B)<eps)return b.p1;
  return b.p1+(b.p2-b.p1)*(B/A);
}

int contains(Polygon g,Point p){
  int n=g.size();
  bool x=false;
  for(int i=0;i<n;i++){
    Vector a=g[i]-p,b=g[(i+1)%n]-p;
    if(abs(cross(a,b))<eps && dot(a,b)<eps)return 1;
    if(a.y>b.y)swap(a,b);
    if(a.y<eps && eps<b.y && cross(a,b)>eps)x=!x;
  }
  if(x)return 2;
  return 0;
}

bool merge_if_able(Segment &s,Segment t){
  if(!equals(0.0,cross(s.p2-s.p1,t.p2-t.p1)))return false;
  if(abs(ccw(s.p1,t.p1,s.p2))==1)return false;
  if(ccw(s.p1,s.p2,t.p1)==-2 || ccw(t.p1,t.p2,s.p1)==-2)
    return false;
  s=Segment(min(s.p1,t.p1),max(s.p2,t.p2));
  return true;
}

void merge_segments(vector<Segment> &s){
  for(int i=0;i<s.size();i++)
    if(s[i].p2<s[i].p1)swap(s[i].p2,s[i].p1);
  for(int i=0;i<s.size();i++){
    for(int j=i+1;j<s.size();j++){
      if(merge_if_able(s[i],s[j])){
        s[j--]=s.back();
        s.pop_back();
      }
    }
  }
}

typedef vector<vector<int> > Graph;

Graph SegmentArrangement(vector<Segment> v,vector<Point> &ps){
  for(int i=0;i<v.size();i++){
    ps.push_back(v[i].p1);
    ps.push_back(v[i].p2);
    for(int j=i+1;j<v.size();j++){
      if(!intersect(v[i],v[j]))continue;
      ps.push_back(getCrossPointLL(v[i],v[j]));
    }
  }
  Graph g(ps.size());
  for(int i=0;i<v.size();i++){
    vector<pair<double,int> > list;
    for(int j=0;j<ps.size();j++)
      if(ccw(v[i].p1,v[i].p2,ps[j])==0)
        list.push_back(mp(norm(v[i].p1-ps[j]),j));
    sort(list.begin(),list.end());
    for(int j=0;j+1<list.size();j++){
      int a=list[j].s,b=list[j+1].s;
      g[a].push_back(b);
    }
  }
  return g;
}

int n,r;
Polygon p;
Graph g;
vector<Point> ps;

void init(){
  p.clear();
  ps.clear();
}

vector<Point> getS(Point a){
  vector<Point> res;
  res.pb(Point(a.x+r,a.y));
  res.pb(Point(a.x,a.y+r));
  res.pb(Point(a.x-r,a.y));
  res.pb(Point(a.x,a.y-r));
  return res;
}

double getDis(Segment s,Point a){
  double res=abs(s.p1.x-a.x)+abs(s.p1.y-a.y);
  res=min(res,abs(s.p2.x-a.x)+abs(s.p2.y-a.y));
  Segment t;
  t=Segment(Point(-inf,a.y),Point(inf,a.y)); 
  if(intersect(s,t)){
    Point b=getCrossPointLL(s,t);
    res=min(res,abs(a-b));
  }
  t=Segment(Point(a.x,-inf),Point(a.x,inf)); 
  if(intersect(s,t)){
    Point b=getCrossPointLL(s,t);
    res=min(res,abs(a-b));
  }
  return res;
}

bool check(Point a){
  double dis=inf;
  FOR(i,0,n){
    Segment s(p[i],p[(i+1)%n]);
    dis=min(dis,getDis(s,a));
  }
  return equals(dis,r);
}

bool check(Point a,Point b){
  if(!check(a+(b-a)/2.0))return false;
  if(contains(p,a)!=2)return false;
  if(contains(p,b)!=2)return false;
  return true;
}

double solve(){
  double res=0;
  vector<Segment> vs;
  vector<vector<Point> > vp;
  FOR(i,0,n)vp.pb(getS(p[i]));
  FOR(i,0,n){
    vector<Point> a=vp[i],b=vp[(i+1)%n];
    FOR(j,0,4){
      vs.pb(Segment(a[j],b[j]));
      vs.pb(Segment(a[j],a[(j+1)%4]));
      vs.pb(Segment(b[j],b[(j+1)%4]));
    }
  }
  merge_segments(vs);
  g=SegmentArrangement(vs,ps);
  FOR(i,0,g.size()){
    FOR(j,0,g[i].size()){
      Point a=ps[i],b=ps[g[i][j]];
      if(check(a,b))res+=abs(b-a);
    }
  }
  return res;
}

int main()
{
  while(cin>>n>>r && n){
    init();
    FOR(i,0,n){
      int x,y;
      cin>>x>>y;
      p.pb(Point(x,y));
    }
    pd(solve());
  }
  return 0;
}