/*
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;
}
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