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:
// #define _GLIBCXX_DEBUG // for STL debug (optional)
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
using namespace std;

#define debug(...) fprintf(stderr, __VA_ARGS__)
#define int long long int
 
template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}
 
typedef pair<int, int> pii;
typedef long long ll;
 
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const ll INF = 1001001001001001LL;
const ll MOD = 1000000007LL;

int solve() {
    int N; cin >> N;
    if(N == 0) return 0;

    long double pw; cin >> pw;
    vector<long double> vw(N), pf(N), vf(N), th(N);
    for(int i=0; i<N; i++) {
        cin >> vw[i] >> pf[i] >> vf[i] >> th[i];
    }

    auto calc_cost = [&](long double x) {
        long double cost_sum = x * pw;
        for(int i=0; i<N; i++) {
            long double f = (th[i] - x * vw[i]) / vf[i];
            cost_sum += pf[i] * max(f, (long double)0.0);
        }
        return cost_sum;
    };

    long double ub = 1e10, lb = 0;
    for(int i=0; i<200; i++) {
        long double l = (lb * 2 + ub) / 3;
        long double r = (lb + 2 * ub) / 3;
        if(calc_cost(l) < calc_cost(r)) {
            ub = r;
        }
        else {
            lb = l;
        }
    }

    printf("%.12Lf\n", calc_cost(ub));
    return 1;
}

signed main() {
    while(solve());
    return 0;
}