Problem: The hobbits Frodo and Sam are carrying the One Ring to Mordor. In order not to be spotted by orcs, they decided to go through the mountains.

The mountain relief can be represented as a polyline with n points (x_i, y_i), numbered from 1 to n (x_i < x_{i + 1} for 1 ≤ i ≤ n - 1). Hobbits start their journey at the point (x_1, y_1) and should reach the point (x_n, y_n) to complete their mission.

The problem is that there is a tower with the Eye of Sauron, which watches them. The tower is located at the point (x_n, y_n) and has the height H, so the Eye is located at the point (x_n, y_n + H). In order to complete the mission successfully, the hobbits have to wear cloaks all the time when the Sauron Eye can see them, i. e. when there is a direct line from the Eye to the hobbits which is not intersected by the relief.

The hobbits are low, so their height can be considered negligibly small, but still positive, so when a direct line from the Sauron Eye to the hobbits only touches the relief, the Eye can see them.

<image> The Sauron Eye can't see hobbits when they are in the left position, but can see them when they are in the right position.

The hobbits do not like to wear cloaks, so they wear them only when they can be spotted by the Eye. Your task is to calculate the total distance the hobbits have to walk while wearing cloaks.

Input

The first line of the input contains two integers n and H (2 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ H ≤ 10^4) — the number of vertices in polyline and the tower height.

The next n lines contain two integers x_i, y_i each (0 ≤ x_i ≤ 4 ⋅ 10^5; 0 ≤ y_i ≤ 10^4) — the coordinates of the polyline vertices. It is guaranteed that x_i < x_{i + 1} for 1 ≤ i ≤ n - 1.

Output

Print one real number — the total distance the hobbits have to walk while wearing cloaks. Your answer will be considered correct if its absolute or relative error does not exceed 10^{-6} — formally, if your answer is a, and the jury's answer is b, your answer will be accepted if (|a - b|)/(max(1, b)) ≤ 10^{-6}.

Examples

Input


6 10
10 40
20 10
25 30
30 15
50 15
65 30


Output


70.4034587602


Input


9 5
0 0
5 10
15 10
20 0
25 11
30 0
35 10
50 10
60 5


Output


27.2787986124


Input


2 10000
0 10000
400000 0


Output


400124.9804748512

Solution (in CPP):

Solution:
#include<bits/stdc++.h> 
using namespace std;
namespace Sakurajima_Mai{
	#define ms(a) memset(a,0,sizeof(a))
	#define repi(i,a,b) for(int i=a,bbb=b;i<=bbb;++i)//attention reg int or reg ll ?
	#define repd(i,a,b) for(int i=a,bbb=b;i>=bbb;--i)
	#define reps(s) for(int i=head[s],v=e[i].to;i;i=e[i].nxt,v=e[i].to)
	#define ce(i,r) i==r?'\n':' '
	#define pb push_back 
	#define all(x) x.begin(),x.end()
	#define gmn(a,b) a=min(a,b)
	#define gmx(a,b) a=max(a,b)
	#define fi first
	#define se second   					  
	typedef long long ll;
	typedef unsigned long long ull;
	typedef double db;
	const int infi=2e9;//infi??,????inf????int
	const ll infl=4e18;
	inline ll ceil_div(ll a,ll b){ return (a+b-1)/b; }
	//std::mt19937 rnd(time(0));//std::mt19937_64 rnd(time(0));
}
using namespace Sakurajima_Mai;
namespace Fast_Read{
	inline int qi(){
	    int f=0,fu=1; char c=getchar();
	    while(c<'0'||c>'9'){ if(c=='-')fu=-1; c=getchar(); }
	    while(c>='0'&&c<='9'){ f=(f<<3)+(f<<1)+c-48; c=getchar(); }
	    return f*fu;
	}
	inline ll ql(){
		ll f=0;int fu=1; char c=getchar();
	    while(c<'0'||c>'9'){ if(c=='-')fu=-1; c=getchar(); }
	    while(c>='0'&&c<='9'){ f=(f<<3)+(f<<1)+c-48; c=getchar(); }
	    return f*fu;
	}
	inline db qd(){
	    char c=getchar();int flag=1;double ans=0;
	    while((!(c>='0'&&c<='9'))&&c!='-') c=getchar();
	    if(c=='-') flag=-1,c=getchar();
	    while(c>='0'&&c<='9') ans=ans*10+(c^48),c=getchar();
	    if(c=='.'){c=getchar();for(int Bit=10;c>='0'&&c<='9';Bit=(Bit<<3)+(Bit<<1)) ans+=(double)(c^48)*1.0/Bit,c=getchar();}
	    return ans*flag;
	}
}
namespace Read{
	#define si(a) scanf("%d",&a)
	#define sl(a) scanf("%lld",&a)
	#define sd(a) scanf("%lf",&a)
	#define ss(a) scanf("%s",a)
	#define rai(x,a,b) repi(i,a,b) x[i]=qi()
	#define ral(x,a,b) repi(i,a,b) x[i]=ql()
}
namespace Out{
	#define pi(x) printf("%d",x)
	#define pl(x) printf("%lld",x)
	#define ps(x) printf("%s",x)
	#define pc(x) printf("%c",x)
	#define pe() puts("")
}
namespace DeBug{
	#define MARK false
	#define DB if(MARK) 
	#define pr(x) cout<<#x<<": "<<x<<endl
	#define pra(x,a,b) cout<<#x<<": "<<endl; \
			repi(i,a,b) cout<<x[i]<<" ";	 \
			puts("");   
	#define FR(a) freopen(a,"r",stdin)
	#define FO(a) freopen(a,"w",stdout)
}
using namespace Fast_Read;
using namespace Read;
using namespace Out;
using namespace DeBug;
const int MAX_N=2e5+5;
const db pi=acos(-1.0);
const db eps=1e-10;
inline int sgn(db x){ return fabs(x)<eps?0:(x<0?-1:1); }
struct Point{
	db x,y;
	Point(){}
	Point(db _x,db _y){x=_x,y=_y;}
	void input(){sd(x),sd(y);} 
	bool operator ==(Point b)const{return sgn(x-b.x)==0&&sgn(y-b.y)==0;}
	Point operator +(const Point &b)const{return Point(x+b.x,y+b.y);}
	Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}
	Point operator /(const db b)const{return Point(x/b,y/b);}
    Point operator *(const db b)const{return Point(x*b,y*b);}
    db operator ^(const Point &b)const{return x*b.y-y*b.x;}//²æ³Ë
	db operator *(const Point &b)const{return x*b.x+y*b.y;}//µã³Ë
	db dis(Point p){ return sqrt((x-p.x)*(x-p.x)+(y-p.y)*(y-p.y)); }//hypot(x-p.x,y-p.y)
}p[MAX_N];
struct Line{
	Point s,e;
	Line(){}
	Line(Point _s,Point _e){ s=_s,e=_e; }
	pair<int,Point> operator &(const Line &b)const
	{
		Point res=s;
		if(sgn((s-e)^(b.s-b.e))==0){
			if(sgn((s-b.e)^(b.s-b.e))==0) return make_pair(0,res);
			else return make_pair(1,res);
		}
		db t=((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
		res.x+=(e.x-s.x)*t,res.y+=(e.y-s.y)*t;
		return make_pair(2,res);
	}
};
bool Seg_inter_Line(Line l1,Line l2)//??l2?????????l1??
{  
	return sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e))<=0;
} 
db Get_K(int i,int j){ return (p[j].y-p[i].y)/(p[j].x-p[i].x); }
int main()
{
	int n=qi(),h=qi();
	repi(i,1,n) p[i].input();
	p[n+1]=p[n]; p[n+1].y+=h;
	db ans=p[n].dis(p[n-1]),k=Get_K(n+1,n-1);
	Line sight(p[n+1],p[n-1]);
	bool see=true;
	repd(i,n-1,2){
		db nk=Get_K(i-1,i);
		if(see){
			if(sgn(nk-k)>0) see=false;
			else ans+=p[i].dis(p[i-1]),k=Get_K(i-1,n+1),sight=Line(p[n+1],p[i-1]);
		}
		else{
			if(sgn(nk-k)>0) continue;
			else{
				Line road(p[i],p[i-1]);
				if(!Seg_inter_Line(sight,road)) continue;
				see=true;
				ans+=(sight&road).se.dis(p[i-1]);
				k=Get_K(i-1,n+1),sight=Line(p[n+1],p[i-1]);
			}			
		}
	}	
	printf("%.10lf\n",ans);
	return 0;
}