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>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
using namespace std;

template<typename T>
ostream& operator << (ostream& os, vector<T>& v){
	rep(i,v.size()){ os << v[i] << (i == v.size() - 1 ? "" : " "); } return os;
}
template<typename T>
istream& operator >> (istream& is, vector<T>& v){
	for(T& x: v){ is >> x; } return is;
}

int main(){
	int n;
	while(cin >> n, n){
		double ww; // weight water
		cin >> ww;

		vector<double> vw(n), wf(n), vf(n), th(n);
		rep(i,n){
			cin >> vw[i] >> wf[i] >> vf[i] >> th[i];
		}

		auto C = [&](double w){
			double res = 0;
			rep(i,n){
				double a = (th[i] - w * vw[i]) / vf[i];
				if(a < 0) res += 0;
				else res += a * wf[i];
			}
			//cout << fixed << setprecision(8)<< w << ' ' << res + w * ww << endl;
			return res + w * ww;
		};


		double right = 1e15, left = 0;
		rep(i,200){
			double llr = (left * 2 + right) / 3;
			double rll = (left + right * 2) / 3;
			if(C(llr) > C(rll)){
				left = llr;
			}else{
				right = rll;
			}
		}

		cout << fixed << setprecision(8)  << C(left) << endl;

	}
}