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.util.*;
import java.io.*;
import java.math.*;
 
 
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
 
public class Main{
    static final long mod=1000000007;
    
    public  static void main(String[] args)   throws Exception, IOException{
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);
        // int n=sc.nextInt();
        // char c[][] = new char[h][w];
        // char c[]=sc.nextString().toCharArray();
        // for(int i=0; i<n; i++) {a[i]=sc.nextInt();}
        int n=sc.nextInt();
        int m=sc.nextInt();
        int start = sc.nextInt() - 1, goal = sc.nextInt() - 1;

        long d[] = new long[n];
        long g[] = new long[n];
        fill(d, 1L<<62);
        fill(g, 1L<<62);
        d[start] = 0;
        g[goal] = 0;
        
        ArrayList<Long> lise[] = new ArrayList[n];
        ArrayList<Long> liss[] = new ArrayList[n];
        long sf=1L<<20;
        for (int i=0; i<n; i++)lise[i] =  new ArrayList<Long>();
        for (int i=0; i<n; i++)liss[i] =  new ArrayList<Long>();
        for (int i=0; i<m; i++) {
            int a=sc.nextInt()-1,
                b=sc.nextInt()-1;
              long  q=sc.nextLong(),
                    w=sc.nextLong();
                lise[a].add( q*sf + b);
                lise[b].add( q*sf + a);
                liss[a].add( w*sf + b);
                liss[b].add( w*sf + a);
        }

        PriorityQueue<P> que = new PriorityQueue<P>();
        que.add(new P(start, d[start]));

        while(!que.isEmpty()) {
            P p = que.poll();   
            int cr = p.id;
            // db(cr);
            if(d[cr] < p.d) continue;
            long or, es; int to;
            for( int t=0; t<lise[cr].size(); t++ ){
                or = lise[cr].get(t);
                to = (int)( or & (sf-1));
                es = or>>20;
                if( d[to] > d[cr] + es) {
                    d[to] = d[cr] + es;
                    que.add(new P(to, d[to]));
                }
            }
        }
        // db(d);

        // que.clear();
        // que.add(new P(goal, g[goal]));
        // while(!que.isEmpty()) {
        //     P p = que.poll();
        //     int cr = p.id;
        //     // db(cr);
        //     if(g[cr] < p.d) continue;
        //     long or, es; int to;
        //     for( int t=0; t<liss[cr].size(); t++ ){
        //         or = liss[cr].get(t);
        //         to = (int)( or & (sf-1));
        //         es = or>>20;
        //         // db(to,es);
        //         if( g[to] > g[cr] + es) {
        //             g[to] = g[cr] + es;
        //             que.add(new P(to, g[to]));
        //         }
        //     }
        // }
        // // db(g);
        // P p[] = new P[n];
        // for( int t=0; t<n; t++ ){
        //     p[t] = new P(t, d[t]+g[t] );
        // }
        // sort(p);
        // long M=1000000000000000L;
        // int cr=p[0].id, x=0;

        // for( int t=0; t<n; t++ ){
          
        //         while(p[x].id<=t-1)x++;
        //         // db(x);
        //         out.println(M-min(p[x].d, d[goal]));
        //     // db(p[t].id,p[t].d);
        // }

        // out.println();
        out.flush();
    }
    

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }
     
    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }
    
    static boolean bit(long x,int k){
        // weather k-th bit (from right) be one or zero
        return  ( 0 < ( (x>>k) & 1 )  )  ? true:false;
    }    
}

class P implements Comparable<P>{
        long d;
        int id;
        P(int  id, long d) {
            this.id=id;
            this.d=d;
        } 
          
        public int compareTo(P p){
            return (d - p.d)>0?1:-1;//des
        } 
    }

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString() throws IOException
    {
        while( st==null || !st.hasMoreTokens() )
            st = new StringTokenizer(x.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException
    {
        return Integer.parseInt(nextString());
    }
    public long nextLong() throws IOException
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}