#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 100000 + 5;

int N, M, L, V, v[MAX_N];
double t1[MAX_N], t2[MAX_N], suf[MAX_N];
pair <double, int> event[MAX_N << 1], tmp1[MAX_N], tmp2[MAX_N];

double solve(int n, double t) {
    double cur = 0;
    int sz1 = 0, sz2 = 0;
    static double t1[MAX_N], t2[MAX_N];
    for (int i = n; i > 0; i --) {
        t1[i] = (L - v[i] * t) / (double)v[i];
        t2[i] = 2.0 * (L - v[i] * t) / (v[i] + V);
        cur += t1[i];
        tmp1[++ sz1] = pair((L - v[i] * t) / (v[i] + V), i);
        if (v[i] > V) tmp2[++ sz2] = pair(0.5 * (L - v[i] * t) / v[i], -i);
    }
    M = sz1 + sz2;
    for (int i = 1, j = 1, k = 1; i <= M; i ++) {
        if (k > sz2 || (j <= sz1 && tmp1[j].first < tmp2[k].first)) event[i] = tmp1[j ++];
        else event[i] = tmp2[k ++];
    }
    double ans = cur;
    int cnt = 0;
    for (int i = 1; i <= M; i ++) {
        auto [t0, pos] = event[i];
        if (pos < 0) {
            cnt ++;
            cur -= t1[-pos];
        }else {
            if (v[pos] > V) {
                cnt --;
            }else {
                cur -= t1[pos];
            }
            cur += t2[pos];
        }
        ans = min(ans, cur + 2.0 * cnt * t0);
    }
    return ans + t * n;
}

int main() {
    scanf("%d%d%d", &N, &L, &V);
    for (int i = 1; i <= N; i ++) {
        scanf("%d", v + i);
        t1[i] = (double)L / (double)v[i];
        t2[i] = 2.0 * L / (v[i] + V);
    }
    for (int i = N; i > 0; i --) suf[i] = suf[i + 1] + t1[i];
    int l = 1, r = N - 1;
    double ans = solve(N, 0);
    while (r - l > 1) {
        int mid1 = (l * 2 + r) / 3, mid2 = (l + r * 2 + 1) / 3;
        double res1 = solve(mid1, t1[mid1 + 1]) + suf[mid1 + 1];
        double res2 = solve(mid2, t1[mid2 + 1]) + suf[mid2 + 1];
        if (res1 < res2) {
            r = mid2 - 1;
            ans = min(ans, res1);
        }else {
            l = mid1 + 1;
            ans = min(ans, res2);
        }
    }
    for (int i = l; i <= r; i ++) ans = min(ans, solve(i, t1[i + 1]) + suf[i + 1]);
    printf("%.10lf\n", ans / N);
    return 0;
}