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

typedef pair<double,double> P;

#define repl(i,a,b) for(int (i)=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define dbg(x) cout<<#x<<"="<<x<<endl

#define EPS 1e-6

struct line{
  double wi,ai,bi;
  int typ;
};
bool operator<(const line& a,const line& b){
  return a.wi < b.wi;
}

int N;
int pw;
int vw[100010],pf[100010],vf[100010],th[100010];

int main(){
  while(1){
    cin>>N;
    if(N==0)break;
    cin>>pw;
    rep(i,N)cin>>vw[i]>>pf[i]>>vf[i]>>th[i];

    double a=0,b=0;
    vector<line> ls;
    rep(i,N){

      double ai=-(double)pf[i]*((double)vw[i]/(double)vf[i]);
      double bi=(double)pf[i]*((double)th[i]/(double)vf[i]);
      
      if(ai>EPS){ // pos
        double wi=-bi/ai;
        if(wi>EPS){
          ls.push_back((line){wi,ai,bi,+1});
        }else{
          a+=ai;
          b+=bi;
        }
      }else if(ai<-EPS){ // neg
        double wi=-bi/ai;
        if(wi>EPS){
          ls.push_back((line){wi,ai,bi,-1});
          a+=ai;
          b+=bi;
        }else{
      
        }
      }else{
        b+=max(0.0,bi);
      }
    }
    sort(all(ls));
    double res=b;
    rep(i,ls.size()){
      double crtw=ls[i].wi;
      res=min(res,crtw*((double)pw+a)+b);
      if(ls[i].typ==+1){
        a+=ls[i].ai;
        b+=ls[i].bi;
      }else{
        a-=ls[i].ai;
        b-=ls[i].bi;
      }
    }
    printf("%.10f\n", res);
  }

  return 0;
}