import heapq

def dijkstra(n, graph, start):
    INF = float('inf')
    dist = [INF] * (n + 1)
    dist[start] = 0
    heap = []
    heapq.heappush(heap, (0, start))
    while heap:
        current_dist, u = heapq.heappop(heap)
        if current_dist > dist[u]:
            continue
        for v, c in graph[u]:
            if dist[v] > dist[u] + c:
                dist[v] = dist[u] + c
                heapq.heappush(heap, (dist[v], v))
    return dist

n, m, P, Q, T = map(int, input().split())
graph = [[] for _ in range(n+1)]
for _ in range(m):
    a, b, c = map(int, input().split())
    graph[a].append((b, c))
    graph[b].append((a, c))

# Compute shortest paths from 1, P, Q
d1 = dijkstra(n, graph, 1)
dp = dijkstra(n, graph, P)
dq = dijkstra(n, graph, Q)

max_time = -1

# Check Scenario A: 1->P->Q->1 and 1->Q->P->1
time1 = d1[P] + dp[Q] + dq[1]
if time1 <= T:
    max_time = max(max_time, T)
time2 = d1[Q] + dq[P] + dp[1]
if time2 <= T:
    max_time = max(max_time, T)

# Check Scenario B: Split at some city X
for X in range(1, n+1):
    max_pq = max(2 * dp[X], 2 * dq[X])
    total_time = 2 * d1[X] + max_pq
    if total_time <= T:
        candidate = T - max_pq
        if candidate > max_time:
            max_time = candidate

# Check Scenario D: Separate visits to P and Q
time_p = 2 * d1[P]
time_q = 2 * d1[Q]
max_pq_separate = max(time_p, time_q)
if max_pq_separate <= T:
    candidate = T - max_pq_separate
    max_time = max(max_time, candidate)

print(max_time if max_time != -1 else -1)