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 <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

struct data{
	double t;
	int v;
	data(int vv=0,double tt=0){
		v=vv;
		t=tt;
	}
	bool operator<(const data &d1)const{
		return t<d1.t;
	}
};

int n;
double pw;
double vw[100001],pf[100001],vf[100001],th[100001];
data query[100001];

int main(void){
	while(1){
		scanf("%d",&n);
		if(n==0)break;
		scanf("%lf",&pw);
		double sum=0.0;
		double dif=0.0;
		double res=0.0;
		int size=0;
		for(int i=0;i<n;i++){
			scanf("%lf %lf %lf %lf",&vw[i],&pf[i],&vf[i],&th[i]);
			if(th[i]>0){
				res+=th[i]/vf[i]*pf[i];
				sum+=th[i]/vf[i]*pf[i];
				if(vw[i]>0){
					query[size++]=data(i,th[i]/vw[i]);
				}
				dif+=vw[i]/vf[i]*pf[i];
			}else{
				if(vw[i]<0){
					query[size++]=data(i,-th[i]/-vw[i]);
				}
			}
		}
		sort(query,query+size);
		double prev=0.0;
		for(int i=0;i<size;i++){
			sum-=dif*(query[i].t-prev);
			res=min(res,max(sum,0.0)+query[i].t*pw);
			if(vw[query[i].v]<0)dif+=vw[query[i].v]/vf[query[i].v]*pf[query[i].v];
			if(vw[query[i].v]>0)dif-=vw[query[i].v]/vf[query[i].v]*pf[query[i].v];
			prev=query[i].t;
		}
		printf("%.10f\n",res);
	}
	return 0;
}