'''
倒着想，假设B一开始有100块钱，求倒着走，怎么才能走到A的时候钱最少
每条边的权值可以抽象为 w = 100 / (100-z)
其实求的是 w1 * w2 * w3 ..... 的最小值
如果能求log(w1*w2*w3)的最小值，也就是log(w1) + log(w2) .... 的最小值
也就等价得到了w1 * w2 * w3 ..... 的最小值
其实就转换为了单源最短路问题，边的权值做了一个对数变换
'''

from collections import deque

class SPFA:

    # start_node 为起始点，edges是边的三元组(节点1， 节点2， 边权重) is_directed表示是否是有向图
    def __init__(self, start_node, edges, is_directed, max_node_num):
        self.edges = edges[::]
        self.start_node = start_node
        self.is_directed = is_directed
        self.max_node_num = max_node_num

    # 获取最短路径长度的列表[(节点1，长度1), (节点2, 长度2) .....]
    def getMinPathLen(self):
        return self.getMinInfo()

    def getMinInfo(self):
        link = {}
        dis = [0x7fffffff] * (self.max_node_num+1)
        dis[self.start_node] = 0

        for a, b, w in self.edges:
            if not self.is_directed:
                # 无向图检查是否有负边
                if w < 0:
                    return None

                if a not in link:
                    link[a] = []
                if b not in link:
                    link[b] = []

                link[a].append((b, w))
                link[b].append((a, w))

            else:
                if a not in link:
                    link[a] = []
                link[a].append((b, w))


        updated_nodes = deque()
        updated_nodes.append(self.start_node)
        in_que = [0] * (self.max_node_num + 1)
        in_que[self.start_node] = 1

        # 进行V次迭代，第V次检查是否是无解情况
        iter_times = 0
        while iter_times < self.max_node_num:
            update_flag = False

            # 利用上一轮迭代距离变小的点进行松弛操作
            node_num = len(updated_nodes)
            for _ in range(node_num):
                a = updated_nodes.popleft()
                in_que[a] = 0
                if a not in link:
                    continue

                for b, w in link[a]:
                    new_dis = 0x7fffffff if dis[a] == 0x7fffffff else dis[a] + w
                    if new_dis < dis[b]:
                        dis[b] = new_dis
                        update_flag = True

                        if in_que[b] == 0:
                            in_que[b] = 1
                            updated_nodes.append(b)

            iter_times += 1

            if iter_times == self.max_node_num:
                if update_flag:
                    return None

            if not update_flag:
                break

        return [[node, dis[node]] for node in range(1, self.max_node_num+1)]


import math

n, m = map(int, input().split())
edges = []

for i in range(m):
    a, b, z = map(int, input().split())
    w = math.log2(100 / (100-z))
    edges.append((a, b, w))

A, B = map(int, input().split())

algo = SPFA(B, edges, is_directed=False, max_node_num=n)
ans = algo.getMinPathLen()
for node, w in ans:
    if node == A:
        print('{:.8f}'.format(100 * (2**w)))