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: Wrong Answer)

import java.math.BigInteger;
import java.util.*;

public class Main {
	int INF = 1 << 30;
	
	class State implements Comparable<State>{
		int now;
		long cost;
		public State(int now, long cost) {
			this.now = now;
			this.cost = cost;
		}
		public int compareTo(State o) {
			return (int) (this.cost - o.cost);
		}
		@Override
		public String toString() {
			return "State [now=" + now + ", cost=" + cost + "]";
		}
		
	}
	
	public class Edge implements Comparable<Edge>{
		int from,to, cost;
		public Edge(int from, int to, int cost) {
			this.from = from; this.to = to; this.cost = cost;
		}
		public int compareTo(Edge o) {
			if(this.cost < o.cost)return -1;
			if(this.cost > o.cost)return 1;
			return 0;
		}
		@Override
		public String toString() {
			return "Edge [from=" + from + ", to=" + to + ", cost=" + cost + "]";
		}
		
	}
	
	private void doit(){
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			int n = sc.nextInt();
			int m = sc.nextInt();
			int s = sc.nextInt() -1;
			int t = sc.nextInt() -1;
			ArrayList<ArrayList<Edge>>pass1 = new ArrayList<>();
			ArrayList<ArrayList<Edge>>pass2 = new ArrayList<>();
			for(int i = 0; i < n; i++){
				pass1.add(new ArrayList<>());
				pass2.add(new ArrayList<>());
			}
			
			for(int i = 0; i < m; i++){
				int a = sc.nextInt() - 1;
				int b = sc.nextInt() - 1;
				int c = sc.nextInt();
				int d = sc.nextInt();
				pass1.get(a).add(new Edge(a, b, c));
				pass1.get(b).add(new Edge(b, a, c));
				pass2.get(a).add(new Edge(a, b, d));
				pass2.get(b).add(new Edge(b, a, d));
			}
			
			PriorityQueue<State> open = new PriorityQueue<>();
			long [] close1 = new long[n];
			Arrays.fill(close1, INF);
			close1[s] = 0;
			open.add(new State(s, 0));
			while(! open.isEmpty()){
				State now = open.poll();
				for(int i = 0; i < pass1.get(now.now).size(); i++){
					long nextcost = pass1.get(now.now).get(i).cost;
					int nextindx = pass1.get(now.now).get(i).to;
					if(nextcost == INF) continue;
					nextcost = now.cost + nextcost;
					if(nextcost >= close1[nextindx]) continue;
					close1[nextindx] = nextcost;
					open.add(new State(nextindx, nextcost));
				}
			}
//			debug(close1);
			
			PriorityQueue<State> open2 = new PriorityQueue<>();
			long [] close2 = new long[n];
			Arrays.fill(close2, INF);
			close2[t] = 0;
			open2.add(new State(t, 0));
			while(! open2.isEmpty()){
				State now = open2.poll();
				for(int i = 0; i < pass2.get(now.now).size(); i++){
					int next = pass2.get(now.now).get(i).cost;
					int nextTo = pass2.get(now.now).get(i).to;
					if(next == INF) continue;
					long nextcost = now.cost + next;
					if(nextcost >= close2[nextTo]) continue;
					close2[nextTo] = nextcost;
					open2.add(new State(nextTo, nextcost));
				}
			}
//			debug(close2);
			
			ArrayList<Long> list = new ArrayList<>();
			long max = 0;
			for(int i = n-1; i >= 0; i--){
				long now = 1000000000000000L - close1[i] - close2[i];
				max = Math.max(max, now);
				list.add(max);
			}
			
			StringBuilder sb = new StringBuilder();
			for(int i = list.size()-1; i >= 0; i--){
				sb.append(list.get(i) + "\n");
			}
			System.out.print(sb.toString());
		}
	}

	private void debug(Object... o) {
		System.out.println("debug = " + Arrays.deepToString(o));
	}

	public static void main(String[] args) {
		new Main().doit();
	}

}