// Original C++ solution (slower):
#include<iostream>

#include<cstdio>

#include<cmath>

#include<queue>

#include<vector>

#include<cstring>

#define INF 1e10

using namespace std;

struct edge{

  int to,limit;

  double dist;

};

struct Point{

  int speed,prev,v;

  double cost;

  Point(){}

  Point(int speed_,int prev_,int v_,double cost_)

    :speed(speed_),prev(prev_),v(v_),cost(cost_){}

  bool operator>(const Point &a)const{

    return cost > a.cost;

  }

};

vector<edge> G[100];

double d[40][40][40];//speed,pos,prev



double dikstra(int s,int g){

  int ds[3]={-1,0,1};

  priority_queue<Point,vector<Point>,greater<Point> > que;

  

  for(int i=0;i<40;i++)

    for(int j=0;j<40;j++)

      for(int k=0;k<40;k++)

	d[i][j][k]=INF;

  

  d[0][s][0]=0;

  Point tmp(0,0,s,0);

  que.push(tmp);

  

  while(!que.empty()){

    Point p=que.top();que.pop();

    if(d[p.speed][p.v][p.prev]<p.cost)continue;

    if(p.speed==1 && p.v==g)return p.cost;



    for(int i=0;i<G[p.v].size();i++){

	edge e=G[p.v][i];

      for(int j=0;j<3;j++){

	int ns=p.speed+ds[j];

	if(1<=ns && ns<= e.limit &&

	   e.to!=p.prev && 

	   d[ns][e.to][p.v]>p.cost+e.dist/ns){



	  Point tmp(ns,p.v,e.to,p.cost+e.dist/ns);

	  que.push(tmp);

	  d[ns][e.to][p.v]=p.cost+e.dist/ns;

	  //cout << "d[" << ns << "][" << e.to << "][" << p.v << "]:" << d[ns][e.to][p.v] << endl;

	}

      }

    }

  }

  return -1;

}



int main(){

  int n,m,s,g;

  while(cin>>n>>m,n){

    for(int i=0;i<100;i++)G[i].clear();

    cin>>s>>g;

    for(int i=0;i<m;i++){

      int x,y,d,c;

      cin>>x>>y>>d>>c;

      G[x].push_back((edge){y,c,d});

      G[y].push_back((edge){x,c,d});

    }

    double ans=dikstra(s,g);

    if(ans==-1)cout << "unreachable" << endl;

    else printf("%.5lf\n",ans);

  }

  return 0;

}

// Optimized solution to the same problem (faster):
//#define DEBUG

#include <bits/stdc++.h>

using namespace std;

/*{{{*/  //template

#define REP(i,n) for(int i=0;i<n;i++)

#define rep(i,n) for(int i=0;i<n;i++)

#define INF 1<<29

#define LINF LLONG_MAX/3

#define MP make_pair

#define PB push_back

#define EB emplace_back

#define ALL(v) (v).begin(),(v).end()

#define all(v) ALL(v)

#define debug(x) cerr<<#x<<":"<<x<<endl

#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl

#define CININIT cin.tie(0),ios::sync_with_stdio(false)

struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;

struct Double{ double d; explicit Double(double x) : d(x){} };

ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }

template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }

template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int,int> pii;

typedef vector<int> vi;

typedef vector<vi> vvi;

ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }

 

constexpr ll mod = 1e9+7;

const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};

/*}}}*/

 

// (time,speed,node,pre_node)

using P = pair<pair<double,int>,pair<int,int>>;

 

 

P makeP(double t,int s,int n,int np){

    return make_pair(make_pair(t,s),make_pair(n,np));

}

 

void solve(int n,int m){

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

    start--;

    goal--;

    vector<vector<int>> g(n,vector<int>(n,INF));

    vector<vector<int>> lim(n,vector<int>(n,0));

    rep(i,m){

        int x,y,d,c;

        cin>>x>>y>>d>>c;

        x--;

        y--;

        g[x][y] = d;

        g[y][x] = d;

        lim[x][y] = c;

        lim[y][x] = c;

    }

 

    double visited[32][32][32];

    rep(i,32) rep(j,32) rep(k,32) visited[i][j][k]=DBL_MAX/10;

    visited[start][0][start] = 0;

 

    priority_queue<P,vector<P>,greater<P>> que;

    que.push(makeP(0,0,start,start));

    while(!que.empty()){

        auto p = que.top();

        que.pop();

        double ti = p.first.first;

        int speed = p.first.second;

        int node = p.second.first;

        int pre = p.second.second;

 

#ifdef DEBUG

        cerr << "time : " << ti << " "  << "node : " << node+1 << " " << "speed " << speed << endl;

#endif

        if(node==goal and speed==1){

            cout << Double(ti) << endl;

            return;

        }

 

        if(ti > visited[node][speed][pre]) continue;

 

        for(int next_node=0;next_node<n;next_node++){

            if(next_node==node or next_node==pre or g[node][next_node]==INF) continue;

            int d=g[node][next_node];

            int c=lim[node][next_node];

            for(int nspeed=speed-1;nspeed<=speed+1;nspeed++){

                if(nspeed<=0 or nspeed>c) continue;

                double nt = ti + (double)d/(double)nspeed;

                if(nt < visited[next_node][nspeed][node]){

#ifdef DEBUG

                    cerr << "\ttime " << nt << " " << "node " << next_node+1 << " speed " << nspeed << endl;

#endif

                    visited[next_node][nspeed][node] = nt;

                    que.push(makeP(nt,nspeed,next_node,node));

                }

            }

        }

    }

    cout << "unreachable" << endl;

}

 

int main(){

    int n,m;

    while(cin>>n>>m){

        if(n==0 && m==0) break;

        solve(n,m);

    }

}

// Fastest known solution for this problem: