/*
    算法描述：
        本题的dist数组表示从原点到其他点还剩原本钱的百分比，因此本题需要最大化d[B]
        初始d[A] = 1，所以更新时采用d[j] = max(d[j], d[t] * g[t][j]);
        g[t][j] = (100.0 - z) / 100，稍微对输入数据进行一下变化
*/

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

using namespace std;

const int N = 2010;

double g[N][N];
int n, m, A, B;
double d[N];
bool st[N];

void dijkstra()
{
    d[A] = 1;
    for (int i = 0; i < n; i ++ )
    {
        int t = -1;
        for (int j = 1; j <= n; j ++ )
            if (!st[j] && (t == -1 || d[t] < d[j]))
                t = j;

        st[t] = true;

        for (int j = 1; j <= n; j ++ )
            d[j] = max(d[j], d[t] * g[t][j]);
    }

    printf("%.8f\n", 100 / d[B]);
}

int main()
{
    scanf("%d%d", &n, &m);
    memset(g, 0x3f, sizeof g);

    while (m -- )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        double w = (100.0 - c) / 100;
        g[a][b] = g[b][a] = max(g[a][b], w);
    }

    scanf("%d%d", &A, &B);
    dijkstra();

    return 0;
}