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;
typedef long long ll;

//dijkstraでsから0=>N-1迄のa_nに関する最短経路問題を解く
//dijkstraで0=>N-1からt迄のb_nに関する最短経路問題を解く

vector<ll> dijkstra(int s, vector<vector<pair<int, int>>> G) {
    const ll INF = numeric_limits<ll> :: max();
    using P = pair<int, int>;
    //vec[i] := sからiへ行くまでの最短距離
    priority_queue<P, vector<P>, greater<P>> pq;
    int N = G.size();
    vector<ll> dist(N, INF);
    dist[s] = 0;
    pq.emplace(P(dist[s], s));
    while(pq.size()) {
        P p = pq.top();
        pq.pop();
        int v = p.second;
        ll d = p.first;
        if (d > dist[v]) {
            continue;
        }
        for (int i = 0; i < G[v].size(); i++) {
            P e = G[v][i];
            int nv = e.first; //nv := vから次に行く点
            ll c = dist[v] + e.second; //c := sからnvまでの暫定最短距離（？）
            if (dist[nv] > c) {
                dist[nv] = c;
                pq.emplace(P(dist[nv], nv));
            }
        }
    }
    return dist;
}

int main() {
    int N, M;
    cin >> N >> M;
    int s, t; cin >> s >> t; s--; t--;
    vector<vector<pair<int, int>>> Ga;
    vector<vector<pair<int, int>>> Gb;
    Ga.resize(N);
    Gb.resize(N);
    vector<ll> yen(N), snu(N);
    for (int i = 0; i < M; i++) {
        int u, v, a, b; cin >> u >> v >> a >> b;
        u--; v--;
        Ga[u].push_back(make_pair(v, a));
        Ga[v].push_back(make_pair(u, a));
        Gb[u].push_back(make_pair(v, b));
        Gb[v].push_back(make_pair(u, b));
    }

    //ダイクストラで配列を生成
    yen = dijkstra(s, Ga); //yen[i] := sから円のみを用いてi番目の都市へ行く最短距離
    snu = dijkstra(t, Gb); //snu[i] := tからすぬーくのみを用いてi番目の都市へ行く最短距離
    //MAX[i]がsからtへ移動した際に所持できるすぬーくの最大値
    vector<ll> MAX(N, 0);
    for (int i = 0; i < N; i++) {
        MAX[i] = 1e15 - yen[i] - snu[i];
    }
    reverse(MAX.begin(), MAX.end());
    vector<ll> ans(N, 0);
    ans[0] = MAX[0];
    for (int i = 1; i < N; i++) {
        ans[i] = max(MAX[i], ans[i-1]);
    }
    reverse(ans.begin(), ans.end());
    for (int i = 0; i < N; i++) {
        cout << ans[i] << endl;
    }
}