//2013-10-31 14:17:50    Accepted    2112    1921MS    408K    2388 B    C++    空信高手

#include <iostream>

#include <string>

#include <map>

using namespace std;

#define N 160

#define INF 0x3F3F3F3F

map<string,int> map1;

int f,cost[N][N];

int path[N],vis[N];

/*==================================================*\

 | Dijkstra 数组实现O （N^2 ）

| Dijkstra --- 数组实现( 在此基础上可直接改为STL 的Queue实现)

 | lowcost[] --- beg 到其他点的最近距离

| path[] -- beg为根展开的树，记录父亲结点

\*==================================================*/

void Dijkstra(int lowcost[N],int n,int beg)

{

    int i,j,min;

    memset(vis,,sizeof(vis));

    vis[beg]=;

    for(i=; i<n; i++)

    {

        lowcost[i]=cost[beg][i];

        path[i]=beg;

    }

    lowcost[beg]=;

    path[beg]=-;

    int pre=beg;

    for(i=; i<n; i++)

    {

        min=INF;

        for(j=; j<n; j++)

            //下面的加法可能导致溢出，INF不能取太大

            if(vis[j]==&&lowcost[pre]+cost[pre][j]<lowcost[j])

            {

                lowcost[j]=lowcost[pre]+cost[pre][j];

                path[j]=pre;

            }

        for(j=; j<n; j++)

            if(vis[j]==&&lowcost[j]<min)

            {

                min=lowcost[j];

                pre=j;

            }

        vis[pre]=;

    }

}

void Init()

{

    int i,j;

    for(i=; i<N; i++)

        for(j=; j<N; j++)

            cost[i][j]=INF;

}

int Judge(string str)

{

    pair<map<string,int>::iterator,bool> iter;

    iter = map1.insert(make_pair(str,++f));

    if(iter.second) return f;

    else

    {

        --f;

        return map1[str];

    }

}

int main()

{

    int n,i,j,k,a,b;

    string firstName,lastName;

    int dis;

   // freopen("input.txt","r",stdin);

    while(cin>>n&&n!=-)

    {

        map1.clear();

        int lowcost[N];

        Init();

        f=;

        cin>>firstName>>lastName;

        a=Judge(firstName);b=Judge(lastName);//a是起点，b是终点

        for(i=; i<n; i++)

        {

            cin>>firstName>>lastName>>dis;

            j=Judge(firstName);k=Judge(lastName);

            cost[j-][k-]=cost[k-][j-]=dis;

        }

        Dijkstra(lowcost,f,a-);

        if(lowcost[b-]<INF)cout<<lowcost[b-]<<endl;

        else cout<<"-1"<<endl;

//        for(map<string,int>::iterator it=map1.begin(); it!=map1.end(); it++)

//            cout<<it->first<<"  "<<it->second<<endl;

    }

    return ;

}