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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

public class Main {
    private static final long INF = 1145141919810893L;
    private static final long INIT = 1000000000000000L;

    private class City {
        long cost = INF;
        boolean fix = false;
        List<Road> roads = new ArrayList<>();
    }

    private class Road {
        int src, dst, cost;

        public Road(int src, int dst, int cost) {
            this.src = src;
            this.dst = dst;
            this.cost = cost;
        }
    }

    public void main(Scanner sc) {
        int n = sc.nextInt();
        int m = sc.nextInt();
        int s = sc.nextInt();
        int t = sc.nextInt();

        City city[] = new City[n];
        City scity[] = new City[n];
        for (int i = 0; i < n; i++) {
            city[i] = new City();
            scity[i] = new City();
        }
        for (int i = 0; i < m; i++) {
            int u = sc.nextInt() - 1;
            int v = sc.nextInt() - 1;
            int a = sc.nextInt();
            int b = sc.nextInt();
            city[u].roads.add(new Road(u, v, a));
            city[v].roads.add(new Road(v, u, a));
            scity[u].roads.add(new Road(u, v, b));
            scity[v].roads.add(new Road(v, u, b));
        }

        long cost[][] = dij(s, t, n, city);
        long scost[][] = dij(s, t, n, scity);
        for (int y = 0; y < n; y++) {
            long min = Long.MAX_VALUE;
            for (int i = y; i < n; i++) {
                min = Math.min(min, cost[s - 1][i] + scost[i][t - 1]);
            }
            System.out.println(INIT - min);
        }
    }

    private long[][] dij(int start, int goal, int n, City city[]) {
        Queue<City> queue = new PriorityQueue<>((o1, o2) -> Long.compare(o1.cost, o2.cost));
        long res[][] = new long[n][n];

        for (int id = 0; id < n; id++) {
            for (int i = 0; i < n; i++) {
                city[i].fix = false;
                city[i].cost = INF;
            }
            queue.clear();

            city[id].cost = 0;
            queue.add(city[id]);
            while (!queue.isEmpty()) {
                City c = queue.poll();
                c.fix = true;
                for (Road r : c.roads) {
                    if (city[r.dst].fix) {
                        continue;
                    }

                    city[r.dst].cost = Math.min(c.cost + r.cost, city[r.dst].cost);
                    queue.add(city[r.dst]);
                }
            }

            for (int i = 0; i < n; i++) {
                res[id][i] = city[i].cost;
            }
        }

        return res;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        new Main().main(sc);
        sc.close();
    }
}