def oi(): return int(input())
def os(): return input()
def mi(): return list(map(int, input().split()))
# import sys
# input = sys.stdin.readline

input_count = 0
N, M, START, GOAL = mi()

G = {i:[] for i in range(N)}
for i in range(M):
    A,B,C = mi()
    
    # Gの第三項に引き継ぎたいものを載せる
    # 何個目～ならi
    # 経路復元なら (A,B)など　行先だけ保存しておけばいいかも
    G[A].append((B, C))
    G[B].append((A, C))


# V: 頂点数
# g[v] = {(w, cost)}:
#     頂点vから遷移可能な頂点(w)とそのコスト(cost)
# r: 始点の頂点
 
from heapq import heappush, heappop
INF = 1<<55


# def trace(s, t, ancestors):
#     route = [t]
#     c = t
#     while True:
#         a = ancestors[c]
#         assert a is not None, 'Failed to trace'
#         route.append(a)
#         if a == s:
#             break
#         c = ancestors[c]
#     route.reverse()
#     return route

# 経路復元の時はコメントアウト部分を解除
def dijkstra(N, G, s):
    dist = [INF] * N
    que = [(0, s)]
    dist[s] = 0
    
    while que:
        c, v = heappop(que)
        if dist[v] < c:
            continue
        for t, cost in G[v]:
            if dist[v] + cost < dist[t]:
                dist[t] = dist[v] + cost
                heappush(que, (dist[t], t))
                
    return dist

d1 = dijkstra(N, G, GOAL)

ret = []
now = START
while now!=GOAL:
    
    mins = INF
    ret.append(now)
    for v,c in G[now]:
        if dist[v] + c == dist[now]:
             mins = min(mins, v)
    now = mins
ret.append(GOAL)

print(*ret)