/*

可以发现， 最优路径上的所有拐点， 基本上都满足一定的性质， 也就是说是在矩形上的拐角处

所以我们可以把他们提出来， 单独判断即可

由于我们提出来的不超过2n + 2个点， 我们将其按照x坐标排序， 这样就只能单方向走

n^2枚举判断两个点之间能不能直接走， 然后连边DAGdp即可

*/

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<queue>

#include<cmath>

#include<iostream>

#define ll long long

#define M 4080

#define mmp make_pair

using namespace std;

const double inf = pow(2, 60);

int read() {

	int nm = 0, f = 1;

	char c = getchar();

	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';

	return nm * f;

}

struct Note {

	int x, y, f;

	bool operator < (const Note &b) const {

		return x == b.x ? y < b.y : x < b.x;

	}

} note[M];

int n, m, L, R, lx[M], ly[M], rx[M], ry[M], bex, bey, edx, edy, tp;

vector<pair<int, double> > to[M];

double f[M], v, ans;

double dis(Note a, Note b) {

	return sqrt(1.0 * (a.x - b.x) * (a.x - b.x) + 1.0 * (a.y - b.y) * (a.y - b.y));

}

int main() {

	n = read();

	for(int i = 1; i <= n; i++) lx[i] = read(), ly[i] = read(), rx[i] = read(), ry[i] = read();

	bex = read(), bey = read(), edx = read(), edy = read();

	if(bex == edx) {

		ans = (bey > edy) ? bey - edy : edy - bey;

	} else {

		if(bex > edx) swap(bex, edx), swap(bey, edy);

		note[++tp] = (Note) {

			bex, bey, 1

		};

		note[++tp] = (Note) {

			edx, edy, 2

		};

		for(int i = 1; i < n; i++) {

			int maxx = max(ly[i], ly[i + 1]), minn = min(ry[i], ry[i + 1]);

			note[++tp] = (Note) {

				lx[i + 1], minn, 3

			};

			note[++tp] = (Note) {

				lx[i + 1], maxx, 4

			};

		}

		sort(note + 1, note + tp + 1);

		for(int i = 1; i <= tp; i++) {

			if(note[i].f == 1) L = i;

			if(note[i].f == 2) R = i;

		}

		for(int i = L; i <= R; i++) {

			double le = -inf, re = inf;

			for(int j = i + 1; j <= R; j++) {

				if(note[j].x == note[i].x) {

					to[j].push_back(mmp(i, note[j].y - note[i].y));

					continue;

				}

				double slp = 1.0 * (note[j].y - note[i].y) / (note[j].x - note[i].x);

				if(slp >= le && slp <= re) to[j].push_back(mmp(i, dis(note[i], note[j])));

				if(note[j].f == 4) le = max(le, slp);

				else re = min(re, slp);

			}

		}

		for(int i = L + 1; i <= R; i++) {

			f[i] = inf;

			for(int j = 0; j < to[i].size(); j++) {

			//	cout << to[i][j].second << "\n";

				f[i] = min(f[i], f[to[i][j].first] + to[i][j].second);

			}

		}

		ans = f[R];

	}

	cin >> v;

	ans /= v;

	printf("%.8lf\n", ans);

	return 0;

}