You are deeply disappointed with the real world, so you have decided to live the rest of your life in the world of MMORPG (Massively Multi-Player Online Role Playing Game). You are no more concerned about the time you spend in the game: all you need is efficiency.

One day, you have to move from one town to another. In this game, some pairs of towns are connected by roads where players can travel. Various monsters will raid players on roads. However, since you are a high-level player, they are nothing but the source of experience points. Every road is bi-directional. A path is represented by a sequence of towns, where consecutive towns are connected by roads.

You are now planning to move to the destination town through the most efficient path. Here, the efficiency of a path is measured by the total experience points you will earn in the path divided by the time needed to travel the path.

Since your object is moving, not training, you choose only a straightforward path. A path is said straightforward if, for any two consecutive towns in the path, the latter is closer to the destination than the former. The distance of two towns is measured by the shortest time needed to move from one town to another.

Write a program to find a path that gives the highest efficiency.



Input

The first line contains a single integer c that indicates the number of cases.

The first line of each test case contains two integers n and m that represent the numbers of towns and roads respectively. The next line contains two integers s and t that denote the starting and destination towns respectively. Then m lines follow. The i-th line contains four integers ui, vi, ei, and ti, where ui and vi are two towns connected by the i-th road, ei is the experience points to be earned, and ti is the time needed to pass the road.

Each town is indicated by the town index number from 0 to (n - 1). The starting and destination towns never coincide. n, m, ei's and ti's are positive and not greater than 1000.

Output

For each case output in a single line, the highest possible efficiency. Print the answer with four decimal digits. The answer may not contain an error greater than 10-4.

Example

Input

2

3 3
0 2
0 2 240 80
0 1 130 60
1 2 260 60

3 3
0 2
0 2 180 60
0 1 130 60
1 2 260 60


Output

3.2500
3.0000
#include <cstdio>
#include <iostream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;

const double EPS = 1.0e-10;
const double INF = DBL_MAX / 1000;

class Edge
{
public:
    int to, cost, experience;
    Edge(){};
    Edge(int to0, int cost0, int experience0){
        to = to0;
        cost = cost0;
        experience = experience0;
    }
};

void shortestPath(const vector<vector<Edge> >& edges, int start, vector<int>& dist)
{
    dist.assign(edges.size(), INT_MAX);
    dist[start] = 0;
    priority_queue<pair<int,int> ,vector<pair<int,int> >, greater<pair<int,int> > > q;
    q.push(make_pair(0, start));

    while(!q.empty()){
        pair<int, int> p = q.top();
        q.pop();
        int v = p.second;
        if(dist[v] < p.first)
            continue;
        for(unsigned i=0; i<edges[v].size(); ++i){
            Edge e = edges[v][i];
            if(dist[v] + e.cost < dist[e.to]){
                dist[e.to] = dist[v] + e.cost;
                q.push(make_pair(dist[e.to], e.to));
            }
        }
    }

    replace(dist.begin(), dist.end(), INT_MAX, -1);
}

int main()
{
    int c;
    cin >> c;

    while(--c >= 0){
        int n, m;
        cin >> n >> m;

        int start, goal;
        cin >> start >> goal;

        vector<vector<Edge> > edges(n);
        for(int i=0; i<m; ++i){
            int u, v, e, t;
            cin >> u >> v >> e >> t;
            edges[u].push_back(Edge(v, t, e));
            edges[v].push_back(Edge(u, t, e));
        }

        vector<int> dist; // ゴールまでの距離
        shortestPath(edges, goal, dist);
        vector<pair<int, int> > d(n);
        for(int i=0; i<n; ++i)
            d[i] = make_pair(dist[i], i);
        sort(d.rbegin(), d.rend());

        // 効率の値effを2分探索
        // Σexp/Σcost >= eff ⇔ Σexp >= eff*Σcost ⇔ Σ(exp - eff*cost) >= 0

        double left = 0.0;
        double right = 1000000.0;
        while(left < right - EPS){
            double eff = (left + right) / 2.0;

            vector<double> dp(n, -INF);
            dp[start] = 0.0;
            for(int i=0; i<n; ++i){
                int a = d[i].second;
                for(unsigned j=0; j<edges[a].size(); ++j){
                    int b = edges[a][j].to;
                    if(dist[b] < dist[a])
                        dp[b] = max(dp[b], dp[a] + (edges[a][j].experience - eff * edges[a][j].cost));
                }
            }

            if(dp[goal] >= 0.0)
                left = eff;
            else
                right = eff;
        }

        printf("%.10f\n", (left + right) / 2.0);
    }

    return 0;
}