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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
class Program{
	static void Main(){
		int[] s=Array.ConvertAll(Console.ReadLine().Split(),int.Parse);
		List<long[]>[] li=new List<long[]>[s[0]+1];
		long[] hy=new long[s[0]+1];
		long[] hs=new long[s[0]+1];
		long[] ans=new long[s[0]];
		for(int i=1;i<=s[0];i++){
			li[i]=new List<long[]>();
			hy[i]=-1;
			hs[i]=-1;
		}
		for(int i=0;i<s[1];i++){
			long[] q=Array.ConvertAll(Console.ReadLine().Split(),long.Parse);
			li[q[0]].Add(new long[]{q[1],q[2],q[3]});
			li[q[1]].Add(new long[]{q[0],q[2],q[3]});
		}

		int m=1;
		heap uy=new heap();
		for(int i=0;i<li[s[2]].Count;i++){uy.ha(li[s[2]][i][1],li[s[2]][i][0]);}
		hy[s[2]]=0;
		while(m<s[0]){
			if(hs[uy.he[0][1]]==-1){
				hy[uy.he[0][1]]=uy.he[0][0];
				for(int i=0;i<li[uy.he[0][1]].Count;i++){
					if(hy[li[uy.he[0][1]][i][0]]==-1){
						uy.ha(li[uy.he[0][1]][i][1]+uy.he[0][0],li[uy.he[0][1]][i][0]);
					}
				}
				m++;
			}
			uy.hr();
		}

		m=1;
		heap us=new heap();
		for(int i=0;i<li[s[3]].Count;i++){us.ha(li[s[3]][i][2],li[s[3]][i][0]);}
		hs[s[3]]=0;
		while(m<s[0]){
			if(hs[us.he[0][1]]==-1){
				hs[us.he[0][1]]=us.he[0][0];
				for(int i=0;i<li[us.he[0][1]].Count;i++){
					if(hs[li[us.he[0][1]][i][0]]==-1){
						us.ha(li[us.he[0][1]][i][2]+us.he[0][0],li[us.he[0][1]][i][0]);
					}
				}
				m++;
			}
			us.hr();
		}
		ans[s[0]-1]=1000000000000000-hy[s[0]]-hs[s[0]];
		for(int i=s[0]-1;i>0;i--){
			ans[i-1]=Math.Max(1000000000000000-hy[i]-hs[i],ans[i]);
		}
		Console.WriteLine(String.Join("\n",ans));
	}
}
public class heap{
	public List<long[]> he=new List<long[]>();
	public void ha(long a,long b){
		he.Add(new long[]{a,b});
		int j=he.Count-1;
		while(he[j][0]<he[(j-1)/2][0]){
			long v=he[j][0];
			he[j][0]=he[(j-1)/2][0];
			he[(j-1)/2][0]=v;
			v=he[j][1];
			he[j][1]=he[(j-1)/2][1];
			he[(j-1)/2][1]=v;
			j=(j-1)/2;
		}
	}
	public void hr(){
		he[0][0]=he[he.Count-1][0];
		he[0][1]=he[he.Count-1][1];
		he.RemoveAt(he.Count-1);
		int j=0;
		while(true){
			if(j*2+2<he.Count){
				if(he[j*2+1][0]<he[j*2+2][0]){j=j*2+1;}
				else{j=j*2+2;}
			}
			else if(j*2+1<he.Count){j=j*2+1;}
			else{break;}
			if(he[j][0]>=he[(j-1)/2][0]){break;}
			long v=he[j][0];
			he[j][0]=he[(j-1)/2][0];
			he[(j-1)/2][0]=v;
			v=he[j][1];
			he[j][1]=he[(j-1)/2][1];
			he[(j-1)/2][1]=v;
		}
	}
}