Problem: 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

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <complex>
#include <vector>
#include <algorithm>
#include <cmath>
#include <array>
using namespace std;

//誤差など
const double EPS = 1e-10;
const double INF = 1e12;
#define EQ(n,m) (abs((n)-(m)) < EPS)
#define X real()
#define Y imag()

//幾何構造
/*
  xy平面上の点は複素数型、
  線分及び直線は要素数2の配列に端点を記憶する
*/
typedef complex<double> P;
typedef vector<P> VP;
struct L : array<P, 2>{
    L(const P& a, const P& b){ at(0)=a; at(1)=b; }
    L(){}
};
namespace std{
    bool operator < (const P& a, const P& b){
        return (a.X!=b.X) ? a.X<b.X : a.Y<b.Y;
    }
    bool operator == (const P& a, const P& b){
        return abs(a-b) < EPS;
    }
}

//基本関数
double dot(P a, P b){
    return (conj(a)*b).X;
}
double cross(P a, P b){
    return (conj(a)*b).Y;
}
int ccw(P a, P b, P c){
    b -= a;
    c -= a;
    if(cross(b,c) > EPS) return +1; //ccw
    if(cross(b,c) < -EPS) return -1; //cw
    if(dot(b,c) < -EPS) return +2; //c-a-b
    if(abs(c)-abs(b) > EPS) return -2; //a-b-c
    return 0; //a-c-b
}

//平行判定
bool isParallel(const P &a, const P &b){
    return abs(cross(a,b)) < EPS;
}
bool isParallel(const L &a, const L &b){
    return isParallel(a[1]-a[0], b[1]-b[0]);
}

//交差判定
bool intersectLS(const L& l, const L& s){
    return cross(l[1]-l[0], s[0]-l[0])*
        cross(l[1]-l[0], s[1]-l[0]) < EPS;
}
bool intersectSS(const L& a, const L& b){
    return ( ccw(a[0],a[1],b[0]) *ccw(a[0],a[1],b[1]) <= 0 ) &&
        ( ccw(b[0],b[1],a[0]) *ccw(b[0],b[1],a[1]) <= 0 );
}
bool intersectSP(const L& s, const P &p){
    return abs(cross(s[0]-p, s[1]-p))<EPS  && dot(s[0]-p, s[1]-p)<EPS;
}
//直線の交点
P crosspointLL(const L &l, const L &m) {
    double A = cross(l[1]-l[0], m[1]-m[0]);
    double B = cross(l[1]-l[0], l[1]-m[0]);
    return m[0] + B/A *(m[1]-m[0]);
}

//多角形の包含判定(1:内側, 0:辺上, -1:外側)
int inPoly(const P &p, const VP &poly){
    int n = poly.size();
    int ret = -1;
    for(int i=0; i<n; i++){
        P a = poly[i]-p;
        P b = poly[(i+1)%n]-p;
        if(a.Y > b.Y) swap(a,b);
        if(intersectSP(L(a,b), P(0,0))) return 0;
        if(a.Y<=0 && b.Y>0 && cross(a,b)<0) ret = 1-ret;
    }
    return ret;
}

//L1ノルム(マンハッタン距離)
double L1distance(const P &a, const P &b){
    return abs(b.X -a.X) +abs(b.Y -a.Y);
}
double L1distanceSP(const L &s, const P &p){
    double ret = INF;
    for(int i=0; i<2; i++){
        L gl(p, p +P(i, 1-i));
        if(!isParallel(s, gl) && intersectLS(gl, s)){
            P cp = crosspointLL(s, L(p, p+P(i,1-i)));
            ret = min(ret, abs(p -cp));
        }
        ret = min(ret, L1distance(p, s[i]));
    }
    return ret;
}

//線分アレンジメント
/*
  線分を重複・交差が無いように分解する
  例えばバツ印は4本の線分になる
*/
vector<L> arrangement(const vector<L> &e){
    vector<VP> cp(e.size());
    for(int i=0; i<(int)e.size(); i++){
        for(int j=i+1; j<(int)e.size(); j++){
            if(!isParallel(e[i], e[j]) && intersectSS(e[i], e[j])){
                P cpij = crosspointLL(e[i], e[j]);
                cp[i].push_back(cpij);
                cp[j].push_back(cpij);
            }
        }
        sort(cp[i].begin(), cp[i].end());
        cp[i].erase(unique(cp[i].begin(), cp[i].end()), cp[i].end());
    }
 
    vector<L> ret;
    for(int i=0; i<(int)e.size(); i++){
        for(int j=0; j<(int)cp[i].size()-1; j++){
            ret.push_back(L(cp[i][j], cp[i][j+1]));
        }
    }
    sort(ret.begin(), ret.end());
    ret.erase(unique(ret.begin(), ret.end()), ret.end());
    return ret;
}


int main(){
    while(true){
        int n,r;
        cin >> n >> r;
        if(n == 0) break;

        //入力受け取り
        VP v(n); //多角形の頂点
        for(int i=0; i<n; i++){
            double x,y;
            cin >> x >> y;
            v[i] = P(x, y);
        }
        
        //各頂点について、上下左右に距離rの点をとって正方形をつくる
        vector<VP> sq(n, VP(4));
        for(int i=0; i<n; i++){
            sq[i][0] = v[i] +P(r, 0);
            sq[i][1] = v[i] +P(0, r);
            sq[i][2] = v[i] +P(-r, 0);
            sq[i][3] = v[i] +P(0, -r);
        }

        /*
          線分リストに、
          (1)もともとの多角形の辺
          (2)各頂点について考えた正方形の辺
          (3)隣り合う頂点の正方形の、対応する頂点を結んだ線分
          を入れる
        */
        vector<L> segs; //線分リスト
        for(int i=0; i<n; i++){
            segs.push_back(L(v[i], v[(i+1)%n]));
            for(int j=0; j<4; j++){
                segs.push_back(L(sq[i][j], sq[(i+1)%n][j]));
                segs.push_back(L(sq[i][j], sq[i][(j+1)%4]));
            }
        }
        //侵食後の海岸線の候補となる線分
        vector<L> cand = arrangement(segs);

        /*
          候補となる線分のうち、最終的に残るのは、
          (1)もともとの多角形よりも内側にあり、
          (2)各辺からマンハッタン距離がr以上
          の点であるので、それらの長さを足し合わせる
         */
        double ans = 0;
        for(int i=0; i<(int)cand.size(); i++){
            bool removed = false;
            for(int j=0; j<n; j++){
                P mid = (cand[i][0] +cand[i][1])/2.0;
                if(L1distanceSP(L(v[j], v[(j+1)%n]), mid) +EPS < r || inPoly(mid, v) == -1){
                    removed = true;
                    break;
                }
            }
            if(!removed){
                ans += abs(cand[i][1] -cand[i][0]);
            }
        }

        //小数点以下10桁で解を出力
        cout << fixed;
        cout << setprecision(10);
        cout << ans << endl;
    }
    return 0;
}