n,m,p,q,t=map(int,input().split())
abc=[list(map(int,input().split())) for _ in range(m)]
p,q=p-1,q-1
g=[[] for _ in range(n)]
for a,b,c in abc:
  a,b=a-1,b-1
  g[a].append([b,c])
  g[b].append([a,c])
inf=float('inf')
from0=[inf]*n
fromp=[inf]*n
fromq=[inf]*n

from0[0]=0
todo=[[0,0]]
from heapq import heappop,heappush
while todo:
  d,v=heappop(todo)
  if from0[v]<d:continue
  for nv,nd in g[v]:
    if from0[nv]>d+nd:
      from0[nv]=d+nd
      heappush(todo,[d+nd,nv])
fromp[p]=0
todo=[[0,p]]
from heapq import heappop,heappush
while todo:
  d,v=heappop(todo)
  if fromp[v]<d:continue
  for nv,nd in g[v]:
    if fromp[nv]>d+nd:
      fromp[nv]=d+nd
      heappush(todo,[d+nd,nv])
fromq[q]=0
todo=[[0,q]]
from heapq import heappop,heappush
while todo:
  d,v=heappop(todo)
  if fromq[v]<d:continue
  for nv,nd in g[v]:
    if fromq[nv]>d+nd:
      fromq[nv]=d+nd
      heappush(todo,[d+nd,nv])
# 別行動しない0->p->q->0
# vで別行動をはじめる。
# 0->v->p->v->0
# 0->v->q->v->0
# vで別行動をはじめるとする。v->p,q最短距離で向かい、再びvで合流するのが最適か
# N<=2000sなのでO(N^2)？ひっかけ？
tmp=from0[p]+fromp[q]+fromq[0]
#print(from0)
#print(fromp)
#print(fromq)
if tmp<=t:
  print(t)
  exit()
ans=-1
for v in range(n):
  tmp=from0[v]*2+max(fromp[v],fromq[v])*2
  if tmp<=t:
    ans=max(ans,t-max(fromp[v],fromq[v])*2)
print(ans)