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>
#define endl '\n'
#define esp ' '
#define fi first
#define se second
#define pb push_back
#define forn(i, a, b) for(int i = a; i < b; i ++)
#define forr(i, a, b) for(int i = a; i >= b; i --)
#define optimizar_io ios_base::sync_with_stdio(0);cin.tie
using namespace std;
using lli = long long int;
using ld = long double;
const ld eps = 1e-9, inf = numeric_limits<ld>::max(), pi = acos(-1);
// For use with integers, just set eps=0 and everything remains the same
bool geq(ld a, ld b){return a-b >= -eps;}     //a >= b
bool leq(ld a, ld b){return b-a >= -eps;}     //a <= b
bool ge(ld a, ld b){return a-b > eps;}        //a > b
bool le(ld a, ld b){return b-a > eps;}        //a < b
bool eq(ld a, ld b){return abs(a-b) <= eps;}  //a == b
bool neq(ld a, ld b){return abs(a-b) > eps;}  //a != b
 
struct point{
	ld x, y;
	point(): x(0), y(0){}
	point(ld x, ld y): x(x), y(y){}
 
	point operator+(const point & p) const{return point(x + p.x, y + p.y);}
	point operator-(const point & p) const{return point(x - p.x, y - p.y);}
	point operator*(const ld & k) const{return point(x * k, y * k);}
	point operator/(const ld & k) const{return point(x / k, y / k);}
 
	point operator+=(const point & p){*this = *this + p; return *this;}
	point operator-=(const point & p){*this = *this - p; return *this;}
	point operator*=(const ld & p){*this = *this * p; return *this;}
	point operator/=(const ld & p){*this = *this / p; return *this;}
 
	point rotate(const ld & a) const{return point(x*cos(a) - y*sin(a), x*sin(a) + y*cos(a));}
	point perp() const{return point(-y, x);}
	ld ang() const{
		ld a = atan2l(y, x); a += le(a, 0) ? 2*pi : 0; return a;
	}
	ld dot(const point & p) const{return x * p.x + y * p.y;}
	ld cross(const point & p) const{return x * p.y - y * p.x;}
	ld norm() const{return x * x + y * y;}
	ld length() const{return sqrtl(x * x + y * y);}
	point unit() const{return (*this) / length();}
 
	bool operator==(const point & p) const{return eq(x, p.x) && eq(y, p.y);}
	bool operator!=(const point & p) const{return !(*this == p);}
	bool operator<(const point & p) const{return le(x, p.x) || (eq(x, p.x) && le(y, p.y));}
	bool operator>(const point & p) const{return ge(x, p.x) || (eq(x, p.x) && ge(y, p.y));}
	bool half(const point & p) const{return le(p.cross(*this), 0) || (eq(p.cross(*this), 0) && le(p.dot(*this), 0));}
};
 
istream &operator>>(istream &is, point & p){return is >> p.x >> p.y;}
ostream &operator<<(ostream &os, const point & p){return os << "(" << p.x << ", " << p.y << ")";}
 
int sgn(ld x){
	if(ge(x, 0)) return 1;
	if(le(x, 0)) return -1;
	return 0;
}
 
point intersectLines(const point & a1, const point & v1, const point & a2, const point & v2){
	//lines a1+tv1, a2+tv2
	//assuming that they intersect
	ld det = v1.cross(v2);
	return a1 + v1 * ((a2 - a1).cross(v2) / det);
}


lli crossP(pair<lli,lli> a, pair<lli,lli> b){
	return a.fi * b.se - a.se * b.fi;
}

pair<lli,lli> resta(pair<lli,lli> &a, pair<lli,lli> &b){
	return make_pair(a.fi - b.fi,a.se-b.se);
}

point getP(pair<lli,lli> &a){
	return point((ld)a.fi,(ld)a.se);
}

point intersectLinesInfo( pair<lli,lli>  a1,  pair<lli,lli> v1,  pair<lli,lli>  a2,  pair<lli,lli>  v2, int &hay){
	//lines a1+tv1 and a2+tv2
	lli det = crossP(v1,v2);
	if(eq(det, 0)){
		pair<lli,lli> aux = resta(a2,a1);
		if(crossP(aux,v1) == 0){
			hay = -1;
			return point(-1,-1); //infinity points
		}else{
			hay =0;
			return point(-1,-1); //single point
		}
	}else{
		hay = 1;
		lli num = crossP(resta(a2,a1),v2);
		ld mult = (ld) num / det;
		return getP(a1) + getP(v1) * mult;
		//return a1 + v1 * ((a2-a1).cross(v2)/det); //no points
	}
}

 
int main(){
    optimizar_io(0);
 
     int n;
     int h;
 
    cin >> n >> h;
    vector<pair<lli,lli>> camino(n);
 
    forn(i, 0, n)
        cin >> camino[i].fi >> camino[i].se;
 	
	pair<lli,lli> ojo = pair<lli,lli>(camino[n-1].fi,camino[n-1].se+h);
 
    bool ve = true;
    ld ans = 0;
    pair<lli,lli> tapa = camino[n-1];
    forr(i, n-2, 0){
        if(crossP(resta(tapa,ojo),resta(camino[i],ojo)) <= 0){
			int hay;
            point intersect = intersectLinesInfo(tapa,resta(tapa,ojo),camino[i],resta(camino[i+1],camino[i]),hay);
            if(hay == -1){
                intersect = getP(camino[i+1]);
            }
            ans += (intersect - getP(camino[i])).length();            
            tapa = camino[i];
        } 
    }
 
    cout << setprecision(10) <<fixed << ans << endl ; 
 
    return 0;
}