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.Arrays;
import java.util.LinkedList;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {

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

	Scanner sc = new Scanner(System.in);

	void run() {

		int n = sc.nextInt();
		int m = sc.nextInt();
		int s = sc.nextInt()-1;
		int t = sc.nextInt()-1;
		
		List<E> map[] = new List[n];
		for(int i = 0; i < n ;i++){
			map[i] = new LinkedList<E>();
		}
		
		for(int i = 0; i < m ;i++){
			E e = new E();
			e.from = sc.nextInt()-1;
			e.to = sc.nextInt()-1;
			e.a = sc.nextInt();
			e.b = sc.nextInt();
			
			map[e.from].add(e);
			map[e.to].add(e);
		}
		
		
		long[] minE = new long[n];
		Arrays.fill(minE, Long.MAX_VALUE);
		{
			PriorityQueue<N> pq = new PriorityQueue<N>((x,y)-> Long.compare(x.cost, y.cost) );
			N init = new N();
			init.pos = s;
			init.cost  =0;
			pq.add(init);
			boolean visited[] = new boolean[n];
			for(;!pq.isEmpty();){
				N now = pq.poll();
				
				if(visited[now.pos]){
					continue;
				}
				visited[now.pos] = true;
				minE[now.pos] = now.cost;
				
				for(E e: map[now.pos]){
					
					int npos = e.from ^ e.to ^ now.pos;
					long ncos = now.cost + e.a;

					if(ncos < minE[npos]){
						minE[npos] = ncos;
						N next = new N();
						next.pos = npos;
						next.cost = ncos;
						pq.add(next);
					}
				}				
			}
		}
		long[] minS = new long[n];
		Arrays.fill(minS, Long.MAX_VALUE);
		{
			PriorityQueue<N> pq = new PriorityQueue<N>((x,y)-> Long.compare(x.cost, y.cost) );
			N init = new N();
			init.pos = t;
			init.cost  =0;
			pq.add(init);
			boolean visited[] = new boolean[n];
			for(;!pq.isEmpty();){
				N now = pq.poll();
				
				if(visited[now.pos]){
					continue;
				}
				visited[now.pos] = true;
				minS[now.pos] = now.cost;
				
				for(E e: map[now.pos]){
					
					int npos = e.from ^ e.to ^ now.pos;
					long ncos = now.cost + e.b;

					if(ncos < minS[npos]){
						minS[npos] = ncos;
						N next = new N();
						next.pos = npos;
						next.cost = ncos;
						pq.add(next);
					}
				}				
			}
		}
		
		long[] ansrev = new long[n];
		ansrev[0] = minE[n-1]+minS[n-1];
		for(int i = 1; i < n ;i++){
			ansrev[i] = Math.min(ansrev[i-1], minE[n-1-i]+minS[n-1-i]);
		}
		
		for(int i = n-1; i >=0; i --){
			System.out.println(1000000000000000L-ansrev[i]);
		}
		
	}
	
	class E{
		int from;
		int to;
		int a;
		int b;
	}
	
	class N{
		int pos;
		long cost;
	}
}