//intersections交叉,相交,十字路口  indices指数,编号，索引（index）
// source起点  In case如果,万一  identical完全相同的  format格式

//being the total number of streets intersections on a map, and the number of streets, respectively.
//分别为地图上的街道交叉口总数和街道数量。

//In case the shortest and the fastest paths are identical, print them in one line in the format:
//如果最短路径和最快路径相同，请按以下格式打印在一行中：

#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>

using namespace std;

//找一个最短路和一个最快的路
//如果最短路线不唯一，则输出用时最短的那条路线（保证唯一）。
//如果最快路线不唯一，则输出经过路口最少的那条路线（保证唯一）。
//并记录路线方案

const int N = 510;

int n,m,S,T;
int d1[N][N],d2[N][N];
int dist1[N],dist2[N],pre[N];
bool st[N];

//返回值，距离和路线pair<int,string>
pair<int,string> dijkstra(int d1[][N],int d2[][N],int type)
{
    memset(dist1,0x3f,sizeof dist1);
    memset(dist2,0x3f,sizeof dist2);
    memset(st,0,sizeof st);

    dist1[S] = dist2[S] = 0;

    for(int i = 0; i<n; i++)
    {
        int t = -1;
        for(int j = 0; j<n; j++)
            if(!st[j] &&(t==-1 || dist1[t] > dist1[j]))
                t = j;
        st[t] = true;
        for(int j = 0; j<n; j++)
        {
            int w;
            if(type == 0) w = d2[t][j];
            else w = 1;

            if(dist1[j] > dist1[t] + d1[t][j])
            {
                dist1[j] = dist1[t] + d1[t][j];
                dist2[j] = dist2[t] + w;
                pre[j] = t;
            }
            else if(dist1[j] == dist1[t] + d1[t][j])
            {
                if(dist2[j] > dist2[t] + w)
                {
                    dist2[j] = dist2[t] + w;
                    pre[j] = t;
                }
            }
        }
    }
    vector<int> path;
    for(int i = T; i!=S; i = pre[i]) path.push_back(i);

    pair<int,string> res;
    res.first = dist1[T];
    res.second = to_string(S);
    for(int i = path.size()-1; i>=0; i--)
        res.second += " -> "+to_string(path[i]);
    return res;
}

int main()
{
    cin>>n>>m;
    memset(d1,0x3f,sizeof d1);
    memset(d2,0x3f,sizeof d2);
    while(m--)
    {
        int a,b,t,c,d;
        cin>>a>>b>>t>>c>>d;
        d1[a][b] = min(d1[a][b],c);
        d2[a][b] = min(d2[a][b],d);
        if(!t)//双行道
        {
            d1[b][a] = min(d1[b][a],c);
            d2[b][a] = min(d2[b][a],d);
        }
    }

    cin>>S>>T;

    auto A = dijkstra(d1,d2,0);
    auto B = dijkstra(d2,d1,1);

    if(A.second != B.second)
    {
        printf("Distance = %d: %s\n",A.first,A.second.c_str());
        printf("Time = %d: %s\n",B.first,B.second.c_str());
    }
    else printf("Distance = %d; Time = %d: %s\n",A.first,B.first,A.second.c_str());
    return 0;
}