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.ArrayList;
import java.util.Scanner;
import java.util.TreeMap;



public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int s = sc.nextInt()-1;
		int t = sc.nextInt()-1;

		ArrayList<ArrayList<Edge>> graph = new ArrayList<ArrayList<Edge>>();

		for (int i=0;i<n;i++) {
			graph.add(new ArrayList<Edge>());
		}

		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();
			Edge e = new Edge();
			e.from=u;
			e.to=v;
			e.yen=a;
			e.snuke=b;
			Edge f = new Edge();
			f.from=v;
			f.to=u;
			f.yen=a;
			f.snuke=b;

			graph.get(u).add(e);
			graph.get(v).add(f);
		}
		long [] yen = new long[n];
		long [] snuke = new long[n];
		TreeMap<Long,Integer> map = new TreeMap<Long, Integer>();
		for(int i=0;i<n;i++) {
			if (i != s) {
				yen[i]=(long)Math.pow(10, 14);

			} else {
				yen[i]=0;
			}
			map.put(yen[i]*n-Long.MAX_VALUE+i, i);
		}


		while (!map.isEmpty()) {
			int node = map.firstEntry().getValue();
			map.remove(map.firstEntry().getKey(), node);
			for (Edge e:graph.get(node)) {
				int toNode = e.to;
				if (yen[toNode] > yen[node]+e.yen && map.containsKey(yen[toNode]*n-Long.MAX_VALUE+toNode)) {
					map.remove(yen[toNode]*n-Long.MAX_VALUE+toNode, toNode);
					yen[toNode] = yen[node]+e.yen;
					map.put(yen[toNode]*n-Long.MAX_VALUE+toNode, toNode);
				}
			}
		}


		map = new TreeMap<Long, Integer>();
		for(int i=0;i<n;i++) {
			if (i != t) {
				snuke[i]=(long)Math.pow(10, 14);

			} else {
				snuke[i]=0;
			}
			map.put(snuke[i]*n-Long.MAX_VALUE+i, i);
		}

		while (!map.isEmpty()) {
			int node = map.firstEntry().getValue();
			map.remove(map.firstEntry().getKey(), node);
			for (Edge e:graph.get(node)) {
				int toNode = e.to;
				if (snuke[toNode] > snuke[node]+e.snuke && map.containsKey(snuke[toNode]*n-Long.MAX_VALUE+toNode)) {
					map.remove(snuke[toNode]*n-Long.MAX_VALUE+toNode, toNode);
					snuke[toNode] = snuke[node]+e.snuke;
					map.put(snuke[toNode]*n-Long.MAX_VALUE+toNode, toNode);
				}
			}
		}

		long min = Long.MAX_VALUE;
		long[] ans = new long[n];
		for (int i=n-1;i>=0;i--) {
			min = Math.min(yen[i]+snuke[i], min);
			ans[i] = min;
		}
		for (int i=0;i<n;i++) {
			System.out.println((long)Math.pow(10, 15)-ans[i]);
		}

	}



}


class Edge {
	int from;
	int to;
	long yen;
	long snuke;
}