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<stdio.h>
#include<algorithm>
using namespace std;
pair<double,pair<int,int> > ev[110000];
int c[110000];int d[110000];
int e[110000];int f[110000];
int main(){
	int a;
	while(scanf("%d",&a),a){
		int b;scanf("%d",&b);
		double sum=0;
		int sz=0;
		for(int i=0;i<a;i++){
			int p,q,r,s;scanf("%d%d%d%d",&p,&q,&r,&s);
			c[i]=p;d[i]=q;e[i]=r;f[i]=s;
			if(s>0&&p>0){
				ev[sz++]=make_pair((double)s/p,make_pair(i,0));
			}
			if(s<=0&&p<0){
				ev[sz++]=make_pair((double)s/p,make_pair(i,1));
			}
		}
		std::sort(ev,ev+sz);
		double ret=0;
		double now=0;
		sum=b;
		for(int i=0;i<a;i++){
			if(f[i]>0){
				now+=(double)d[i]*f[i]/e[i];
				sum-=(double)c[i]*d[i]/e[i];
			}
		}
		ret=now;
		double last=0;
		for(int i=0;i<sz;i++){
		//	printf("%f %f %f\n",now,sum,ret);
			now+=(ev[i].first-last)*sum;
			last=ev[i].first;
			ret=min(ret,now);
			int at=ev[i].second.first;
			if(ev[i].second.second){
				sum-=(double)c[at]*d[at]/e[at];
			}else{
				sum+=(double)c[at]*d[at]/e[at];
			}
		
		}
		printf("%f\n",ret);
	}
}