/*
假设给到B手上100时,A最少花费 
从终点B倒推 到A的最小距离
B→C  dist[C] * (100-w[i])/100 = 100(dist[B])
     dist[C] = 100(dist[B])*100/(100-w[i])
I→J  dist[J] * (100-w[i])/100 = dist[I]
     dist[J] = dist[I]*100/(100-w[i])
*/

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

const int N=2010,M=2000010;

int h[N],ne[M],e[M],w[M],idx;
double dist[N];
int n,m,s,t;
bool st[N];
void add(int a,int b,int c)
{
    e[idx] = b,w[idx]=c,ne[idx] = h[a],h[a]=idx++;
}

double spfa()
{
    memset(dist,0x7f,sizeof dist);//注意double初始化0x3f = 0.0004767
    dist[t] = 100;
    queue<int> q;
    q.push(t);
    st[t]=true;
    while(q.size())
    {
        int ver = q.front();
        q.pop();
        st[ver] = false;
        for(int i=h[ver];i!=-1;i=ne[i])
        {
            int j=e[i];
            if(dist[j]>dist[ver]*100/(100.0-w[i]))
            {
                dist[j]=dist[ver]*100/(100.0-w[i]);
                if(!st[j])
                {
                    st[j]=true;
                    q.push(j); 
                }
            }
        }
    }
    return dist[s];//我们最终结果在dist[s]上
}
dijkstra+堆优化
double dijkstra()
{
    memset(dist,0x7f,sizeof dist);
    priority_queue<PII,vector<PII>,greater<PII>> heap;
    heap.push({100.0,t});

    while(heap.size())
    {
        auto u = heap.top();
        heap.pop();
        double d = u.first;//dist[ver]
        int ver = u.second;
        //顺序又错了 在遍历j之前判断ver 而不是判断j
        if(st[ver]) continue;
        st[ver]=true;

        for(int i=h[ver];i!=-1;i=ne[i])
        {
            int j=e[i];
            if(dist[j]>d*100/(100.0-w[i]))
            {
                dist[j]=d*100/(100.0-w[i]);
                heap.push({dist[j],j});
            }
        }
    }
    return dist[s];
}

int main()
{
    memset(h,-1,sizeof h);
    cin >> n >> m;
    for(int i=0;i<m;i++)
    {
        int a,b,c;
        cin >> a >> b >> c;
        add(a,b,c),add(b,a,c);
    }
    cin >> s >> t;
    printf("%.8lf\n",spfa());
    //printf("%.8lf\n",dijkstra());
    return 0;
}