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.*;
import java.io.*;
class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte(){
        if(ptr < buflen){
            return true;
        }else{
            ptr = 0;
            try{
                buflen = in.read(buffer);
            }catch(IOException e){
                e.printStackTrace();
            }
            if(buflen <=0){
                return false;
            }
        }
        return true;
    }

    private int readByte(){
        if(hasNextByte())return buffer[ptr++];
        else return -1;
    }

    private static boolean isPrintableChar(int c){
        return 33<=c && c<=126;
    }
    public boolean hasNext(){
        while(hasNextByte() && !isPrintableChar(buffer[ptr]))ptr++;
        return hasNextByte();
    }

    public String next(){
        if(!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)){
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong(){
        if(!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        if(b < '0' || '9' < b){
            throw new NumberFormatException();
        }
        while(true){
            if('0' <= b && b<='9'){
                n*=10;
                n+=b-'0';
            }else if(b==-1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt(){
        long nl = nextLong();
        if(nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDoutble(){return Double.parseDouble(next());}
}
public class Main {
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int n = fs.nextInt();
        int m = fs.nextInt();
        int s = fs.nextInt()-1;
        int t = fs.nextInt()-1;
        List<List<Road>> graph = new ArrayList<>();

        for(int i=0;i<n;i++)graph.add(new LinkedList<>());

        int from,to;
        long a,b;
        for(int i=0;i<m;i++){
            from = fs.nextInt()-1;
            to = fs.nextInt()-1;
            a = fs.nextLong();
            b = fs.nextLong();
            graph.get(from).add(new Road(to,a,b));
            graph.get(to).add(new Road(from,a,b));
        }
        long[] cost = new long[n];
        long inf = 1000000000000000L;
        Arrays.fill(cost,inf);
        PriorityQueue<Road> queue = new PriorityQueue<>(new Comparator<Road>() {

			@Override
			public int compare(Road o1, Road o2) {
				if(o1.a > o2.a){
                    return 1;
                }else if(o1.a == o2.a){
                    return 0;
                }else{
                    return -1;
                }
			}
        });

        queue.add(new Road(s,0,0));
        while(!queue.isEmpty()){
            Road r = queue.poll();
            if(r.a >= cost[r.to])continue;
            cost[r.to] = r.a;
            for(Road road : graph.get(r.to)){
                if(cost[r.to] + road.a < cost[road.to]){
                    queue.add(new Road(road.to,cost[r.to] + road.a ,0));
                }
            }
        }

        queue = new PriorityQueue<>(new Comparator<Road>() {
			@Override
			public int compare(Road o1, Road o2) {
				if(o1.b > o2.b){
                    return 1;
                }else if(o1.b == o2.b){
                    return 0;
                }else{
                    return -1;
                }
			}
        });
        long[] r_cost = new long[n];
        Arrays.fill(r_cost,inf);
        queue.add(new Road(t,0,0));
        while(!queue.isEmpty()){
            Road r = queue.poll();
            if(r.b >= r_cost[r.to])continue;
            r_cost[r.to] = r.b;
            for(Road road : graph.get(r.to)){
                if(r_cost[r.to] + road.b < r_cost[road.to]){
                    queue.add(new Road(road.to,0,r_cost[r.to] + road.b));
                }
            }
        }
        long[][] total = new long[n][2];

        for(int i=0;i<n;i++){
            total[i][0] = cost[i] + r_cost[i];
            total[i][1] = i+1;
        }
        Arrays.sort(total,new Comparator<long[]>(){
            @Override
            public int compare(long[] o1, long[] o2) {
                if(o1[0] > o2[0]){
                    return 1;
                }else if(o1[0] == o2[0]){
                    return 0;
                }else{
                    return -1;
                }
            }
        });
        PrintWriter out = new PrintWriter(System.out);
        int idx = 0;
        long money = 1000000000000000L;
        for(int i=0;i<n;i++){
            while(i >= total[idx][1])idx++;
            out.println(money-total[idx][0]);
        }
        out.flush();
    }
    
}
class Road{
    int to;
    long a,b;
    Road(int to ,long a,long b){
        this.to = to;
        this.a = a;
        this.b = b;
    }
}