Problem:
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Problem Statement

We have planted $N$ flower seeds, all of which come into different flowers.
We want to make all the flowers come out together.

Each plant has a value called vitality, which is initially zero.
Watering and spreading fertilizers cause changes on it, and the $i$-th plant will come into flower if its vitality is equal to or greater than $\mathit{th}_i$.
Note that $\mathit{th}_i$ may be negative because some flowers require no additional nutrition.

Watering effects on all the plants.
Watering the plants with $W$ liters of water changes the vitality of the $i$-th plant by $W \times \mathit{vw}_i$ for all $i$ ($1 \le i \le n$), and costs $W \times \mathit{pw}$ yen, where $W$ need not be an integer.
$\mathit{vw}_i$ may be negative because some flowers hate water.

We have $N$ kinds of fertilizers, and the $i$-th fertilizer effects only on the $i$-th plant.
Spreading $F_i$ kilograms of the $i$-th fertilizer changes the vitality of the $i$-th plant by $F_i \times \mathit{vf}_i$, and costs $F_i \times \mathit{pf}_i$ yen, where $F_i$ need not be an integer as well.
Each fertilizer is specially made for the corresponding plant, therefore $\mathit{vf}_i$ is guaranteed to be positive.

Of course, we also want to minimize the cost.
Formally, our purpose is described as "to minimize $W \times \mathit{pw} + \sum_{i=1}^{N}(F_i \times \mathit{pf}_i)$ under $W \times \mathit{vw}_i + F_i \times \mathit{vf}_i \ge \mathit{th}_i$, $W \ge 0$, and $F_i \ge 0$ for all $i$ ($1 \le i \le N$)".
Your task is to calculate the minimum cost.

Input

The input consists of multiple datasets.
The number of datasets does not exceed $100$, and the data size of the input does not exceed $20\mathrm{MB}$.
Each dataset is formatted as follows.

$N$
$\mathit{pw}$
$\mathit{vw}_1$ $\mathit{pf}_1$ $\mathit{vf}_1$ $\mathit{th}_1$
:
:
$\mathit{vw}_N$ $\mathit{pf}_N$ $\mathit{vf}_N$ $\mathit{th}_N$

The first line of a dataset contains a single integer $N$, number of flower seeds.
The second line of a dataset contains a single integer $\mathit{pw}$, cost of watering one liter.
Each of the following $N$ lines describes a flower. The $i$-th line contains four integers, $\mathit{vw}_i$, $\mathit{pf}_i$, $\mathit{vf}_i$, and $\mathit{th}_i$, separated by a space.

You can assume that $1 \le N \le 10^5$, $1 \le \mathit{pw} \le 100$, $-100 \le \mathit{vw}_i \le 100$, $1 \le \mathit{pf}_i \le 100$, $1 \le \mathit{vf}_i \le 100$, and $-100 \le \mathit{th}_i \le 100$.

The end of the input is indicated by a line containing a zero.

Output

For each dataset, output a line containing the minimum cost to make all the flowers come out.
The output must have an absolute or relative error at most $10^{-4}$.

Sample Input

3
10
4 3 4 10
5 4 5 20
6 5 6 30
3
7
-4 3 4 -10
5 4 5 20
6 5 6 30
3
1
-4 3 4 -10
-5 4 5 -20
6 5 6 30
3
10
-4 3 4 -10
-5 4 5 -20
-6 5 6 -30
0

Output for the Sample Input

43.5
36
13.5
0

Solution in C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

constexpr double eps = 1e-8;
constexpr double gratio = (std::sqrt(5)+1)/2.0;

template <typename F>
double golden_section_search(F f, double lb, double ub, int cnt) {
    double x1 = (ub-lb)/(gratio+1) + lb,
           x2 = (ub-lb)*gratio/(gratio+1) + lb;
    double v1 = f(x1), v2 = f(x2);
    for(int i=0; i<cnt; ++i) {
        if(v1 < v2) {
            ub = x2;
            x2 = x1;
            v2 = v1;
            x1 = (ub-lb)/(gratio+1) + lb;
            v1 = f(x1);
        } else {
            lb = x1;
            x1 = x2;
            v1 = v2;
            x2 = (ub-lb)*gratio/(gratio+1) + lb;
            v2 = f(x2);
        }
    }
    return (lb+ub)/2;
}

int main() {
    int N;
    while(cin >> N, N) {
        double pw;
        cin >> pw;
        vector<double> vw(N), pf(N), vf(N), th(N);
        double res = 0;
        for(int i=0; i<N; ++i) {
            cin >> vw[i] >> pf[i] >> vf[i] >> th[i];
            res += max(0.0, th[i] / vf[i] * pf[i]);
        }
        auto f = [&](double W) {
            double cost = W * pw;
            for(int i=0; i<N; ++i) {
                cost += max(0.0, (th[i] - vw[i] * W) / vf[i] * pf[i]);
            }
            return cost;
        };
        cout << fixed << setprecision(10) << f(golden_section_search(f, 0, 1e9, 300)) << endl;
    }
}