Problem: Gaku decided to observe the ant's nest as a free study during the summer vacation. The transparent observation case that his grandpa prepared for his grandchildren is very unique and looks like Figure 1.

Ant nest
Figure 1



This case consists of two congruent convex polygons s1 and s2 and some rectangles. Place one of the rectangular faces other than s1 and s2 so that it touches the floor. Gogo, who started observing, noticed that the height of the soil changed depending on how the bottom was selected, even though the amount of soil contained was the same (Fig. 2).

Ant nest
Figure 2



Gogo wanted to know how high it would be depending on how the case was placed. Create a program that outputs the maximum value of the soil height by inputting the shape of the convex polygon s1, the distance d between s1 and s2, and the volume V of the soil. The soil shall level as soon as the bottom of the case is placed on the floor. The s1 shape of the transparent case is given by n vertices on the two-dimensional coordinate plane. These vertices are entered in a counterclockwise order. n is 3 or more and 100 or less, and d and V are integers of 1 or more and 10,000 or less, respectively. In addition, x and y of the coordinates (x, y) of the vertices of the polygon shall be integers of -1,000 or more and 1,000 or less, respectively. V does not exceed the volume of the case.



input

A sequence of multiple datasets is given as input. The end of the input is indicated by three zero lines. Each dataset is as follows.

1st line n d V (integer integer integer; half-width space delimited)
Coordinates of the first vertex on the second line x y (integer integer; half-width space delimited)
Coordinates of the second vertex on the third line
::
n + 1 Coordinates of the nth vertex on the 1st line


The number of datasets does not exceed 20.

output

Outputs the maximum soil height for each input dataset. The answer must not have an error greater than 0.00001.

Example

Input

4 1 1
0 0
1 0
1 2
0 2
0 0 0


Output

1.000000

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
#include<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
using ld =long double;
const ld eps = 1e-9;


/* 幾何の基本 */

#include <complex>

typedef complex<ld> Point;


const ld pi = acos(-1.0);
const ld dtop = pi / 180.;
const ld ptod = 1. / dtop;

namespace std {
	bool operator<(const Point &lhs, const Point &rhs) {
		if (lhs.real() < rhs.real() - eps) return true;
		if (lhs.real() > rhs.real() + eps) return false;
		return lhs.imag() < rhs.imag();
	}
}

// 点の入力
Point input_Point() {
	ld x, y;
	cin >> x >> y;
	return Point(x, y);
}

// 誤差つき等号判定
bool eq(const ld a, const ld b) {
	return (abs(a - b) < eps);
}

// 内積
ld dot(const Point& a, const Point& b) {
	return real(conj(a) * b);
}

// 外積
ld cross(const Point& a, const Point& b) {
	return imag(conj(a) * b);
}

// 直線の定義
class Line {
public:
	Point a, b;
	Line() : a(Point(0, 0)), b(Point(0, 0)) {}
	Line(Point a, Point b) : a(a), b(b) {}
	Point operator[](const int _num)const {
		if (_num == 0)return a;
		else if (_num == 1)return b;
		else {
			assert(false);
			return Point();
		}
	}
};

// 円の定義
class Circle {
public:
	Point p;
	ld r;
	Circle() : p(Point(0, 0)), r(0) {}
	Circle(Point p, ld r) : p(p), r(r) {}
};

// ccw
// 1: a,b,cが反時計周りの順に並ぶ
//-1: a,b,cが時計周りの順に並ぶ
// 2: c,a,bの順に直線に並ぶ
//-2: a,b,cの順に直線に並ぶ
// 0: a,c,bの順に直線に並ぶ
int ccw(const Point& a, const Point &b, const Point &c) {
	const Point nb(b - a);
	const Point nc(c - a);
	if (cross(nb, nc) > eps) return 1;   // a,b,cが反時計周りの順に並ぶ
	if (cross(nb, nc) < -eps) return -1; // a,b,cが時計周りの順に並ぶ
	if (dot(nb, nc) < 0) return 2;       // c,a,bの順に直線に並ぶ
	if (norm(nb) < norm(nc)) return -2;  // a,b,cの順に直線に並ぶ
	return 0;                          // a,c,bの順に直線に並ぶ
}


/* 交差判定 */

// 直線と直線の交差判定
bool isis_ll(const Line& l, const Line& m) {
	return !eq(cross(l.b - l.a, m.b - m.a), 0);
}

// 直線と線分の交差判定
bool isis_ls(const Line& l, const Line& s) {
	return isis_ll(l, s) &&
		(cross(l.b - l.a, s.a - l.a) * cross(l.b - l.a, s.b - l.a) < eps);
}

// 線分と線分の交差判定
bool isis_ss(const Line& s, const Line& t) {
	return ccw(s.a, s.b, t.a) * ccw(s.a, s.b, t.b) <= 0 &&
		ccw(t.a, t.b, s.a) * ccw(t.a, t.b, s.b) <= 0;
}

// 点の直線上判定
bool isis_lp(const Line& l, const Point& p) {
	return (abs(cross(l.b - p, l.a - p)) < eps);
}

// 点の線分上判定
bool isis_sp(const Line& s, const Point& p) {
	return (abs(s.a - p) + abs(s.b - p) - abs(s.b - s.a) < eps);
}

// 垂線の足
Point proj(const Line &l, const Point& p) {
	ld t = dot(p - l.a, l.b - l.a) / norm(l.a - l.b);
	return l.a + t * (l.b - l.a);
}

//線対象の位置にある点
Point reflect(const Line &l, const Point &p) {
	Point pr = proj(l, p);
	return pr * 2.l - p;
}

// 直線と直線の交点
Point is_ll(const Line &s, const Line& t) {
	Point sv = s.b - s.a, tv = t.b - t.a;
	assert(cross(sv, tv) != 0);
	return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}
// 直線と直線の交点
vector<Point> is_ll2(const Line &s, const Line& t) {
	Point sv = s.b - s.a, tv = t.b - t.a;
	if (cross(sv, tv) != 0)return vector<Point>(1, is_ll(s, t));
	else {
		vector<Point>ans;
		for (int k = 0; k < 2; ++k) {
			if (isis_sp(s, t[k]) && find(ans.begin(), ans.end(), t[k]) == ans.end())ans.push_back(t[k]);
			if (isis_sp(t, s[k]) && find(ans.begin(), ans.end(), s[k]) == ans.end())ans.push_back(s[k]);
		}
		return ans;
	}
}
// 線分と線分の交点
//　重なってる部分あるとassert(false)
Point is_ss(const Line &s, const Line& t) {
	if (isis_ss(s, t)) {
		for (int k = 0; k < 2; ++k) {
			for (int l = 0; l < 2; ++l) {
				if (s[k] == t[l])return s[k];
			}
		}
		return is_ll(s, t);
	}
	else {
		//先にisis_ssしてね
		assert(false);
		return Point(0, 0);
	}
}
// 線分と線分の交点
vector<Point> is_ss2(const Line &s, const Line& t) {
	vector<Point> kouho(is_ll2(s, t));
	vector<Point>ans;
	for (auto p : kouho) {
		if (isis_sp(s, p) && isis_sp(t, p))ans.emplace_back(p);
	}
	return ans;
}
// 直線と点の距離
ld dist_lp(const Line& l, const Point& p) {
	return abs(p - proj(l, p));
}

//直線と直線の距離
ld dist_ll(const Line& l, const Line& m) {
	return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}

// 直線と線分の距離
ld dist_ls(const Line& l, const Line& s) {
	return isis_ls(l, s) ? 0 : min(dist_lp(l, s.a), dist_lp(l, s.b));
}

// 線分と点の距離
ld dist_sp(const Line& s, const Point& p) {
	Point r = proj(s, p);
	return isis_sp(s, r) ? abs(r - p) : min(abs(s.a - p), abs(s.b - p));
}

// 線分と線分の距離
ld dist_ss(const Line& s, const Line& t) {
	if (isis_ss(s, t)) return 0;
	return min({ dist_sp(s, t.a), dist_sp(s, t.b), dist_sp(t, s.a), dist_sp(t, s.b) });
}


//直線と直線の二等分線のベクトル
Line line_bisection(const Line &s, const Line &t) {
	const Point laglanju(is_ll(s, t));
	const Point avec = !(abs(laglanju - s[0])<eps) ? s[0] - laglanju : s[1] - laglanju;
	const Point bvec = !(abs(laglanju - t[0])<eps) ? t[0] - laglanju : t[1] - laglanju;

	return Line(laglanju, laglanju + (abs(bvec)*avec + abs(avec)*bvec) / (abs(avec) + abs(bvec)));
}
//点と点の垂直二等分線　aを左に見ながら
Line point_bisection(const Point&a, const Point&b) {
	const Point cen((a + b) / 2.l);
	const Point vec = (b - a)*Point(0, 1);
	return Line(cen, cen + vec);
}




/* 多角形 */

typedef vector<Point> Polygon;



// 面積
ld get_area(const Polygon &p) {
	ld res = 0;
	int n = p.size();
	for (int j = 0; j<n; ++j) res += cross(p[j], p[(j + 1) % n]);
	return res / 2;
}

//多角形の回転方向
bool is_counter_clockwise(const Polygon &poly) {
	ld angle = 0;
	int n = poly.size();
	for(int i=0;i<n;++i) {
		Point a = poly[i], b = poly[(i + 1) % n], c = poly[(i + 2) % n];
		angle += arg((c - b) / (b - a));
	}
	return angle > eps;
}

// 円の内外判定
/*0 => out
1 => on
2 => in*/
int is_in_Polygon(const Polygon &poly, const  Point& p) {
	ld angle = 0;
	int n = poly.size();
	for(int i=0;i<n;++i) {
		Point a = poly[i], b = poly[(i + 1) % n];
		if (isis_sp(Line(a, b), p)) return 1;
		angle += arg((b - p) / (a - p));
	}
	return eq(angle, 0) ? 0 : 2;
}
//円の内外判定2　高速
enum { out, on, in };
int convex_contains(const Polygon &P, const Point &p) {
	const int n = P.size();
	Point g = (P[0] + P[n / 3] + P[2 * n / 3]) / 3.0l; // inner-point
	int a = 0, b = n;
	while (a + 1 < b) { // invariant: c is in fan g-P[a]-P[b]
		int c = (a + b) / 2;
		if (cross(P[a] - g, P[c] - g) > 0) { // angle < 180 deg
			if (cross(P[a] - g, p - g) > 0 && cross(P[c] - g, p - g) < 0) b = c;
			else                                                  a = c;
		}
		else {
			if (cross(P[a] - g, p - g) < 0 && cross(P[c] - g, p - g) > 0) a = c;
			else                                                  b = c;
		}
	}
	b %= n;
	if (cross(P[a] - p, P[b] - p) < 0) return 0;
	if (cross(P[a] - p, P[b] - p) > 0) return 2;
	return 1;
}

// 凸包
//点や線を返すことも有り得るので注意
Polygon convex_hull(vector<Point> ps) {
	int n = ps.size();
	int k = 0;
	sort(ps.begin(), ps.end());
	Polygon ch(2 * n);
	for (int i = 0; i < n; ch[k++] = ps[i++])
		while (k >= 2 && ccw(ch[k - 2], ch[k - 1], ps[i]) <= 0) --k;
	for (int i = n - 2, t = k + 1; i >= 0; ch[k++] = ps[i--])
		while (k >= t && ccw(ch[k - 2], ch[k - 1], ps[i]) <= 0) --k;
	ch.resize(k - 1);
	return ch;
}



//凸カット
vector<Polygon> convex_cut(const Polygon &ps, const Line& l) {
	int n = ps.size();
	Polygon q;
	Polygon r;
	for(int i=0;i<n;++i) {
		Point a = ps[i], b = ps[(i + 1) % n];
		Line m = Line(a, b);
		if (ccw(l.a, l.b, a) != -1) q.push_back(a);
		if (ccw(l.a, l.b, a) != 1) r.push_back(a);
		if (ccw(l.a, l.b, a) * ccw(l.a, l.b, b) < 0 && isis_ll(l, m)) {
			q.push_back(is_ll(l, m));
			r.push_back(is_ll(l, m));
		}
	}
	const vector<Polygon>polys{ q,r };
	return polys;
}


int main()
{
	while (true) {
		int N;cin>>N;
		if(!N)break;
		ld D,V;cin>>D>>V;
		ld S=V/D;

		Polygon poly(N);
		for (int i = 0; i < N; ++i) {
			int x,y;cin>>x>>y;
			poly[i]=Point(x,y);
		}

		ld ans=0;
		for (int i = 0; i < N; ++i) {
			Line l(poly[i],poly[(i+1)%N]);

			Line bis=point_bisection(l.a,l.b);
			Point vec(bis.b-bis.a);

			vec/=abs(vec);

			ld amin=0;
			ld amax=1e6;
			int num=100;
			while (num--) {
				ld amid(amin/2+amax/2);
				Polygon cut_poly=convex_cut(poly,Line(l.a+vec*amid,l.b+vec*amid))[1];

				ld ar=get_area(cut_poly);

				if(ar<S)amin=amid;
				else amax=amid;
			}
			ans=max(ans,amin);
		}
		cout<<setprecision(10)<<fixed<<ans<<endl;

	}
	return 0;
}