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: Accepted)

import java.util.*;

public class Main {
    static Scanner scanner = new Scanner(System.in);

    public static void main(String[]$) {
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        int s = scanner.nextInt() - 1;
        int t = scanner.nextInt() - 1;
        List<Grid>[] yen = new ArrayList[n];
        List<Grid>[] snuuk = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            yen[i] = new ArrayList<>();
            snuuk[i] = new ArrayList<>();
        }

        for (int i = 0; i < m; i++) {
            int u = scanner.nextInt() - 1;
            int v = scanner.nextInt() - 1;
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            yen[u].add(new Grid(v, a));
            yen[v].add(new Grid(u, a));
            snuuk[u].add(new Grid(v, b));
            snuuk[v].add(new Grid(u, b));
        }

        long[] dist1 = dijkstra(yen, s);
        long[] dist2 = dijkstra(snuuk, t);
        long[] ans = new long[n];

        for (int i = n - 1; i >= 0; i--) {
            ans[i] = Math.min(i == n - 1 ? Long.MAX_VALUE : ans[i + 1], dist1[i] + dist2[i]);
        }

        for (long a : ans) {
            System.out.println((long)1e15 - a);
        }
    }

    static long[] dijkstra(List<Grid>[] list, int start) {
        long[] cost = new long[list.length];
        Arrays.fill(cost, 1L << 62);
        cost[start] = 0;

        Queue<Grid> queue = new PriorityQueue<>(Comparator.comparing(grid -> grid.cost));
        queue.add(new Grid(start, 0));

        while (!queue.isEmpty()) {
            Grid grid = queue.poll();
            if (cost[grid.to] >= grid.cost) {
                for (Grid g : list[grid.to]) {
                    if (cost[g.to] > cost[grid.to] + g.cost) {
                        cost[g.to] = cost[grid.to] + g.cost;
                        queue.add(new Grid(g.to, cost[g.to]));
                    }
                }
            }
        }
        return cost;
    }

    static class Grid {
        int to;
        long cost;

        Grid(int to, long cost) {
            this.to = to;
            this.cost = cost;
        }
    }
}