def oi(): return int(input())
def os(): return input()
def mi(): return list(map(int, input().split()))


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
    G[A].append((B, C, f"{A}_{B}"))
    G[B].append((A, C, f"{B}_{A}"))


# 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

# 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
    # 経路復元用
    edge = {i:set([]) for i in range(N)}
    
    while que:
        c, v = heappop(que)
        if dist[v] < c:
            continue
        for t, cost, ind in G[v]:
            if dist[v] + cost < dist[t]:
                dist[t] = dist[v] + cost
                
                # 経路復元用
                edge[t] = set([ind])
                heappush(que, (dist[t], t))
            
            # コストが同じものもどうにかしたいならここ追加
            elif dist[v] + cost == dist[t]:
                edge[t].add(ind)
                heappush(que, (dist[t], t))            
    return edge 

ret = dijkstra(N, G, START)

START = START
GOAL = GOAL

# もし文字列でハッシュ化してたらここで解除
temp = ret.keys()
for k in temp:
    c = []
    for v in ret[k]:
        c.append(tuple(map(int, list(v.split("_")))))
    ret[k] = sorted(c)

def keiro_with_dijkstr(START, GOAL):
    
    def keiro_hukugen(node, old_node):
        keiro_list = []
        for next_node in ret[node]:
            nn = (set(next_node)-set([node])).pop()
            if nn != old_node:
                flg, keiro = keiro_hukugen(nn, node)
                if flg:
                    return flg, keiro + [node]

        if node == START:
            return True, [START]

        return False, keiro_list
    
    return keiro_hukugen(GOAL, -1)



print(*keiro_with_dijkstr(START, GOAL)[1])