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.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map.Entry;
import java.util.PrimitiveIterator;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.TreeMap;
import java.util.function.Supplier;
import java.util.stream.IntStream;
import java.util.stream.Stream;

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

	class Edge{
		int f,t;
		long jpy,snk;
		public Edge(int f,int t,long l,long m){
			super();
			this.f=f;
			this.t=t;
			this.jpy=l;
			this.snk=m;
		}
	}

	int n=gInt(),m=gInt(),src=gInt()-1,to=gInt()-1;
	void solve(){
		HashMap<Integer,ArrayList<Edge>> g=new HashMap<>(n);

		long[] jpy=new long[n],snk=new long[n];

		for(int i:rep(m)){
			int f=gInt()-1,t=gInt()-1,jp=gInt(),sn=gInt();
			g.computeIfAbsent(f,o->new ArrayList<>()).add(new Edge(f,t,jp,sn));
			g.computeIfAbsent(t,o->new ArrayList<>()).add(new Edge(t,f,jp,sn));
		}
		jpy(g,src,jpy);
		snk(g,to,snk);
		TreeMap<Long,Integer>t=new TreeMap<>();
		for(int i:rep(n))
			t.put(jpy[i]+snk[i],i+1);
		Entry<Long,Integer> o=null;
		for(int i:rep(n)) {
			while(true) {
				o=t.firstEntry();
				if(o.getValue()>i)
					break;
				t.pollFirstEntry();
			}
			System.out.println(1000_000_000_000_000l-o.getKey());
		}
	}
	void jpy(HashMap<Integer,ArrayList<Edge>> g,int s,long[] jpy){
		int[] prev=new int[n];
		Arrays.fill(jpy,Long.MAX_VALUE/3);
		Arrays.fill(prev,-1);
		jpy[s]=0;
		PriorityQueue<Edge> q=new PriorityQueue<>(Comparator.comparingLong(o->o.jpy));
		for(q.add(new Edge(-2,s,0,0));!q.isEmpty();){
			Edge e=q.poll();
			if(prev[e.t]!=-1) continue;
			prev[e.t]=e.f;
			for(Edge f:g.get(e.t)){
				if(jpy[f.t]>e.jpy+f.jpy){
					jpy[f.t]=e.jpy+f.jpy;
					q.add(new Edge(f.f,f.t,e.jpy+f.jpy,0));
				}
			}
		}
	}
	void snk(HashMap<Integer,ArrayList<Edge>> g,int s,long[] snk){
		int[] prev=new int[n];
		Arrays.fill(snk,Long.MAX_VALUE/3);
		Arrays.fill(prev,-1);
		snk[s]=0;
		PriorityQueue<Edge> q=new PriorityQueue<>(Comparator.comparingLong(o->o.snk));
		for(q.add(new Edge(-2,s,0,0));!q.isEmpty();){
			Edge e=q.poll();
			if(prev[e.t]!=-1) continue;
			prev[e.t]=e.f;
			for(Edge f:g.get(e.t)){
				if(snk[f.t]>e.snk+f.snk){
					snk[f.t]=e.snk+f.snk;
					q.add(new Edge(f.f,f.t,0,e.snk+f.snk));
				}
			}
		}
	}

	public static void main(String[] A){
		new Main().solve();
	}

	static int gInt(){
		return Integer.parseInt(s.next());
	}
	static long gLong(){
		return Long.parseLong(s.next());
	}
	static double gDouble(){
		return Double.parseDouble(s.next());
	}

	SupplyingIterator<Integer> ints(int n){
		return new SupplyingIterator<>(n,Main::gInt);
	}
	SupplyingIterator<Long> longs(int n){
		return new SupplyingIterator<>(n,Main::gLong);
	}
	SupplyingIterator<Double> doubles(int n){
		return new SupplyingIterator<>(n,Main::gDouble);
	}
	SupplyingIterator<String> strs(int n){
		return new SupplyingIterator<>(n,s::next);
	}

	Range rep(int i){
		return Range.rep(i);
	}
	Range rep(int f,int t,int d){
		return Range.rep(f,t,d);
	}
	Range rep(int f,int t){
		return rep(f,t,1);
	}
	Range rrep(int f,int t){
		return rep(t,f,-1);
	}

	IntStream REP(int v){
		return IntStream.range(0,v);
	}
	IntStream REP(int l,int r){
		return IntStream.rangeClosed(l,r);
	}

	IntStream INTS(int n){
		return IntStream.generate(Main::gInt).limit(n);
	}
	Stream<String> STRS(int n){
		return Stream.generate(s::next).limit(n);
	}

}
class SupplyingIterator<T> implements Iterable<T>,Iterator<T>{
	int			t;
	Supplier<T>	supplier;

	SupplyingIterator(int t,Supplier<T> supplier){
		this.t=t;
		this.supplier=supplier;
	}

	@Override
	public Iterator<T> iterator(){
		return this;
	}

	@Override
	public boolean hasNext(){
		return t>0;
	}

	@Override
	public T next(){
		--t;
		return supplier.get();
	}

}

class Range implements Iterable<Integer>,PrimitiveIterator.OfInt{
	int to,cur,d;

	Range(int from,int to,int d){
		this.cur=from-d;
		this.to=to;
		this.d=d;
	}

	Range(int n){
		this(0,n-1,1);
	}

	@Override
	public Iterator<Integer> iterator(){
		return this;
	}

	@Override
	public boolean hasNext(){
		return cur+d==to||(cur!=to&&(cur<to==cur+d<to));
	}

	@Override
	public int nextInt(){
		return cur+=d;
	}
	static Range rep(int i){
		return new Range(i);
	}
	static Range rep(int f,int t,int d){
		return new Range(f,t,d);
	}
	static Range rep(int f,int t){
		return rep(f,t,1);
	}
	static Range rrep(int f,int t){
		return rep(f,t,-1);
	}
}