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.*;

public class Main {
	static final long all = (long)1e15;
	static int n,m,s,t;
	static int u[],v[],a[],b[],first[],next[];
	static boolean vis[];
	static long yen[],snuuk[],cost[],mincost = all;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		m = in.nextInt();
		s = in.nextInt();
		t = in.nextInt();
		u = new int[m * 2 + 1];
		v = new int[m * 2 + 1];
		a = new int[m * 2 + 1];
		b = new int[m * 2 + 1];
		first = new int[m * 2 + 1];
	    next = new int[m * 2 + 1];
	    vis = new boolean[n + 1];
	    yen = new long[n + 1];
	    snuuk = new long[n + 1];
	    cost = new long[n + 1];
	    Arrays.fill(first, -1);
	    Arrays.fill(yen,all);
	    Arrays.fill(snuuk, all);
	    for(int i = 0;i < m;i ++) {
	    	u[i] = in.nextInt();
	    	v[i] = in.nextInt();
	    	a[i] = in.nextInt();
	    	b[i] = in.nextInt();
	    	u[i + m] = v[i];
	    	v[i + m] = u[i];
	    	a[i + m] = a[i];
	    	b[i + m] = b[i];
	    	next[i] = first[u[i]];
	    	first[u[i]] = i;
	    	next[i + m] = first[u[i + m]];
	    	first[u[i + m]] = i + m;
	    }
	    yen[s] = 0;
	    int head = 0,tail = 0;
	    int queue[] = new int[n];
	    queue[tail ++] = s;
	    vis[s] = true;
	    while(head != tail) {
	    	int k = first[queue[head]];
	    	while(k != -1) {
	    		if(yen[u[k]] + a[k] < yen[v[k]]) {
	    			yen[v[k]] = yen[u[k]] + a[k];
	    			if(vis[v[k]]) {
	    				queue[tail] = v[k];
	    			    vis[v[k]] = true;
	    			    tail = (tail + 1) % n;
	    			}
	    		}
	    		k = next[k];
	    	}
	    	vis[queue[head]] = false;
	    	head = (head + 1) % n;
	    }
	    snuuk[t] = 0;
	    head = tail = 0;
	    queue[tail ++] = t;
	    vis[t] = true;
	    while(head != tail) {
	    	int k = first[queue[head]];
	    	while(k != -1) {
	    		if(snuuk[u[k]] + b[k] < snuuk[v[k]]) {
	    			snuuk[v[k]] = snuuk[u[k]] + b[k];
	    			if(vis[v[k]]) {
	    				queue[tail] = v[k];
	    			    vis[v[k]] = true;
	    			    tail = (tail + 1) % n;
	    			}
	    		}
	    		k = next[k];
	    	}
	    	vis[queue[head]] = false;
	    	head = (head + 1) % n;
	    }
	    for(int i = n;i > 0;i --) {
	    	cost[i] = Math.min(mincost, yen[i] + snuuk[i]);
	    	mincost = Math.min(mincost, cost[i]);
	    	cost[i] = ((all >= cost[i]) ? (all - cost[i]) : 0);
	    }
	    for(int i = 1;i <= n;i ++) {
	    	System.out.println(cost[i]);
	    }
	}
}