#include <bits/stdc++.h>

using namespace std;

using ll = long long;

struct State {
    int at;
    ll cost;
    int prev;
    State(int at, ll cost, int prev) : at(at), cost(cost), prev(prev) {}
    bool operator>(const State& s) const {
        if (cost != s.cost) return cost > s.cost;
        if (prev != s.prev) return prev > s.prev; //最短経路を辞書順最小にする(省略可)
        return at > s.at;
        //return cost > s.cost;
    }
};

struct Edge {
    int to;
    ll cost;
    Edge(int to, ll cost) : to(to), cost(cost) {}
};

using Graph = vector<vector<Edge> >; //隣接リスト

const ll INF = 1e15;
const int NONE = -1;

//sは始点、mincは最短経路のコスト、prevsは最短経路をたどる際の前の頂点
void dijkstra(int s, const Graph& graph, vector<ll>& minc, vector<int>& prevs){
    minc.assign(graph.size(), INF);
    prevs.assign(graph.size(), NONE);

    priority_queue<State, vector<State>, greater<State> > pq;
    pq.emplace(s, 0, NONE);
    minc[s] = 0;

    while(!pq.empty()) {
        State cur = pq.top();
        pq.pop();
        //if (minc[cur.at] < cur.cost) continue; 
        if (minc[cur.at] < INF) continue;

        for(const Edge& e : graph[cur.at]) {
            ll cost = cur.cost + e.cost;
            if (minc[cur.at] == INF) continue;
            //if (minc[e.to] < cost || minc[e.to] == cost && prevs[e.to] <= cur.at) continue;
            minc[e.to] = cost;
            prevs[e.to] = cur.at;
            pq.emplace(e.to, cost, cur.at);
        }
    }
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m, s, g;
    cin >> n >> m >> s >> g;
    Graph graph(n);
    for (int i = 0; i < m; i++) {
        int a, b;
        ll c;
        cin >> a >> b >> c;
        graph[a].emplace_back(b, c);
        graph[b].emplace_back(a, c);
    }

    vector<ll> minc;
    vector<int> prevs;
    dijkstra(s, graph, minc, prevs);

    vector<int> ans;
    int cur = g;
    while (cur != NONE) {
        ans.push_back(cur);
        cur = prevs[cur];
    }
    reverse(ans.begin(), ans.end());
    int sz = ans.size();
    for (int i = 0; i < sz; i++) {
        cout << ans[i] << " \n"[i + 1 == sz];
    }
    return 0;
}