Problem: 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





Example

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

43.5
36
13.5
0

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
using namespace std;
const double INF = 1e18;

int main(){
    cout << fixed;
    cout << setprecision(15);
    while(1){
        int n;
        cin >> n;
        if(n==0) break;
        
        double pw;
        cin >> pw;
        vector<double> vw(n),pf(n),vf(n),th(n);
        vector<double> w(1, 0);
        w.reserve(n+1);
        for(int i=0; i<n; i++){
            cin >> vw[i] >> pf[i] >> vf[i] >> th[i];
            if(th[i]*vw[i] > 0) w.push_back(th[i]/vw[i]);
        }
        sort(w.begin(), w.end());
        
        int lb=0, ub=w.size()-1;
        while(lb+2 < ub){
            int mid[2] = {lb +(ub-lb)/3, lb +(ub-lb)*2/3};
            double cost[2];
            for(int i=0; i<2; i++){
                cost[i] = w[mid[i]]*pw;
                for(int j=0; j<n; j++){
                    if(w[mid[i]]*vw[j] < th[j]){
                        cost[i] += (th[j] -w[mid[i]]*vw[j])/vf[j] *pf[j];
                    }
                }
            }
            if(cost[0] < cost[1]){
                ub = mid[1];
            }else{
                lb = mid[0];
            }
        }

        double ans = INF;
        for(int i=lb; i<=ub; i++){
            double cost = w[i]*pw;
            for(int j=0; j<n; j++){
                if(w[i]*vw[j] < th[j]){
                    cost += (th[j] -w[i]*vw[j])/vf[j] *pf[j];
                }
            }
            ans = min(ans, cost);
        }
        cout << ans << endl;
    }
    return 0;
}