Problem:
Score : 400 points

Problem Statement

Kenkoooo is planning a trip in Republic of Snuke.
In this country, there are n cities and m trains running.
The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally.
Any city can be reached from any city by changing trains.

Two currencies are used in the country: yen and snuuk.
Any train fare can be paid by both yen and snuuk.
The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.

In a city with a money exchange office, you can change 1 yen into 1 snuuk.
However, when you do a money exchange, you have to change all your yen into snuuk.
That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk.
Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.

Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling.
It is acceptable to do the exchange in City s or City t.

Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange.
For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years.
You can assume that the trip finishes within the year.

Constraints

2 \leq n \leq 10^5

1 \leq m \leq 10^5

1 \leq s,t \leq n

s \neq t

1 \leq u_i < v_i \leq n

1 \leq a_i,b_i \leq 10^9

If i\neq j, then u_i \neq u_j or v_i \neq v_j.

Any city can be reached from any city by changing trains.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m

Output

Print n lines.
In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using VI = vector<ll>;
using VV = vector<VI>;
using VS = vector<string>;

// tourist set
template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string& s) {
  return '"' + s + '"';
}

string to_string(const char* s) {
  return to_string((string) s);
}

string to_string(bool b) {
  return (b ? "true" : "false");
}

string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}

template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}

template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}

void debug_out() { cerr << '\n'; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif
// tourist set end

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(i,b) FOR(i, 0, b)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<'\n'
#define p2(s, t) cout << (s) << " " << (t) << '\n'
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << '\n'
#define p_yes() p("YES")
#define p_no() p("NO")
#define SZ(x) ((int)(x).size())
#define SORT(A) sort(ALL(A))
#define RSORT(A) sort(ALL(A), greater<ll>())

void no(){p_no(); exit(0);}
void yes(){p_yes(); exit(0);}

const ll mod = 1e9 + 7;
const ll inf = 1e18;
const double PI = acos(-1);


struct Edge{
    ll to;
    ll cost;
    Edge(ll to, ll cost): to(to), cost(cost) {}
    Edge(){
        to = 0;
        cost = 0;
    }
};

struct Node{
    ll distance;
    ll index;
    Node(ll d, ll i){
        distance = d;
        index = i;
    }
    Node(){}

    bool operator<(const Node &another) const
    {
        return distance < another.distance;
    }
    bool operator>(const Node &another) const
    {
        return distance > another.distance;
    }
};

struct Dijkstra{
    vector<ll> d;
    vector<vector<Edge> > graph;
    vector<bool> done;

    // ノード数を入れる
    void initialize(ll size){
        d.resize(size);
        done.resize(size);
        graph.resize(size);
        reset();
    }

    void reset(){
        ll N = graph.size();
        FOR(i, 0, N){
            d[i] = inf;
            done[i] = false;
        }
    }
    
    ll distance(int i){
        if(d.size()<=i) return -1;
        return d[i];
    }

    void print_graph(){
        FOR(i, 0, graph.size()){
            cout << i << " -> ";

            for(auto edge : graph[i]){
                cout << edge.to << " ";
            }
            cout << endl;
        }
        p("distance");
        FOR(i, 0, graph.size()){
            ll d = distance(i);
            cout << i << " " << d << endl;
        }
    }

    void register_edge(ll a, ll b, ll cost){
        auto edge = Edge(b, cost);
        graph[a].push_back(edge);
    }

    // 双方向
    void register_edge2(ll a, ll b, ll cost){
      register_edge(a, b, cost);
      register_edge(b, a, cost);
    }

    void calc_shortest_path(ll from=0){
        priority_queue<Node, vector<Node>, greater<Node> > que;
        auto node = Node();
        // 始点
        node.index = from;
        node.distance = 0;
        que.push(node);

        while(!que.empty()){
            // 1番distanceが小さいノード
            Node n = que.top();
            que.pop();

            if(done[n.index]){
                continue;
            }
            
            done[n.index] = true;
            d[n.index] = n.distance;

            auto edge_list = graph[n.index];
            for(auto edge : edge_list){   
                // 短くなるノードがあれば
                if(!done[edge.to] && n.distance + edge.cost < d[edge.to]){
                    ll shorter_distance = n.distance + edge.cost;
                    que.push(Node(shorter_distance, edge.to));
                }
            }
        }
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    ll N, M, S, T;
    cin >> N >> M >> S >> T;
    S--;
    T--;

    Dijkstra dij0; dij0.initialize(N); // money
    Dijkstra dij1; dij1.initialize(N); // snuuk

    rep(i, M){
      ll u, v, a, b;
      cin >> u >> v >> a >> b;
      u--;
      v--;

      dij0.register_edge2(u, v, a);
      dij1.register_edge2(u, v, b);
    }
    dij0.calc_shortest_path(S);
    dij1.calc_shortest_path(T);
    


    VI Ans(N);
    rep(i, N){
      ll cost = dij0.distance(i) + dij1.distance(i);
      chmax(Ans[i], cost);
    }

    for(int i=N-1; i>0; i--){
      if(Ans[i-1]>Ans[i]){
        Ans[i-1] = Ans[i];
      }
    }

    debug(Ans);
    rep(i, N){
      ll rest = 1e15 - Ans[i];
      p(rest);
    }
    
    
    return 0;
}