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 {
    static int n;
    static int m;
    static int s;
    static int t;
    static long arigane = (long) 1e15;
    static int[][] grapht;
    static int[][] graphf;
    static long[] tmemo;
    static long[] fmemo;
    static PriorityQueue<Path> paths;
    static long min = Long.MAX_VALUE;
    static List<Long> ret = new ArrayList<>();
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        m = sc.nextInt();
        s = sc.nextInt() - 1;
        t = sc.nextInt() - 1;
        grapht = new int[n][n];
        graphf = new int[n][n];
        tmemo = new long[n];
        fmemo = new long[n];
        for (int i = 0; i < n; i++) {
            tmemo[i] = -1;
            fmemo[i] = -1;
        }
        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();
            grapht[u][v] = a;
            grapht[v][u] = a;
            graphf[u][v] = b;
            graphf[v][u] = b;
        }

        paths = new PriorityQueue<>(Comparator.comparingLong(x -> x.score));
        paths.add(new Path(s, 0));
        while(!paths.isEmpty()) {
            Path c = paths.poll();
            if(tmemo[c.current] != -1) continue;
            tmemo[c.current] = c.score;
            for (int i = 0; i < n; i++) {
                if(grapht[c.current][i] > 0 && tmemo[i] == -1) {
                    Path tmp = new Path(i, c.score + grapht[c.current][i]);
                    paths.add(tmp);
                }
            }
        }

        paths.add(new Path(t, 0));
        while(!paths.isEmpty()) {
            Path c = paths.poll();
            if(fmemo[c.current] != -1) continue;
            fmemo[c.current] = c.score;
            for (int i = 0; i < n; i++) {
                if(graphf[c.current][i] > 0 && fmemo[i] == -1) {
                    Path tmp = new Path(i, c.score + graphf[c.current][i]);
                    paths.add(tmp);
                }
            }
        }
        for (int i = 0; i < n; i++) {
            min = Math.min(min, tmemo[n - 1 - i] + fmemo[n - 1 - i]);
            ret.add(arigane - min);
        }
        Collections.reverse(ret);
        ret.forEach(System.out::println);
    }

    static class Path {
        int current;
        long score;

        Path(int current, long score) {
            this.current = current;
            this.score = score;
        }
    }
}