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: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.IO;
class Program
{
	static void Main(string[] args)
	{
		string[] str = Console.ReadLine().Split();
		long n = long.Parse(str[0]);
		long m = long.Parse(str[1]);
		long s = long.Parse(str[2]);
		long t = long.Parse(str[3]);
		long[,]a = new long[m,4];
		for(var i=0;i<m;i++){
			string[] str2 = Console.ReadLine().Split();
			a[i,0] = long.Parse(str2[0]);
			a[i,1] = long.Parse(str2[1]);
			a[i,2] = long.Parse(str2[2]);
			a[i,3] = long.Parse(str2[3]);
		}
		long[] nd = new long[n];
		long[] ndf = new long[n];
		for(var i=0;i<n;i++){
			nd[i] = 9999999999999999;
		}
		nd[s-1] = 0;
		var qlist = new List<long>();
		qlist.Add(s-1);
		while(true){
			long dmin = 9999999999999999;
			long dmins = -1;
			for(var i=0;i<qlist.Count;i++){
				if(ndf[qlist[i]] == 0){
					if(nd[qlist[i]] <= dmin){
						dmin = nd[qlist[i]];
						dmins = qlist[i];
					}
				}
			}
			if(dmins == -1){
				break;
			}
			ndf[dmins] = 1;
			qlist.Remove(dmins);
			for(var i=0;i<m;i++){
				if(a[i,0]-1==dmins){
					if(ndf[a[i,1]-1] == 0){
						nd[a[i,1]-1] = Math.Min(nd[a[i,1]-1],dmin+a[i,2]);
						qlist.Add(a[i,1]-1);
					}
				} else if(a[i,1]-1==dmins){
					if(ndf[a[i,0]-1] == 0){
						nd[a[i,0]-1] = Math.Min(nd[a[i,0]-1],dmin+a[i,2]);
						qlist.Add(a[i,0]-1);
					}
				}
			}
		}
		long[] nd2 = new long[n];
		long[] ndf2 = new long[n];
		for(var i=0;i<n;i++){
			nd2[i] = 9999999999999999;
		}
		nd2[t-1] = 0;
		var qlist2 = new List<long>();
		qlist2.Add(t-1);
		while(true){
			long dmin = 9999999999999999;
			long dmins = -1;
			for(var i=0;i<qlist2.Count;i++){
				if(ndf2[qlist2[i]] == 0){
					if(nd2[qlist2[i]] <= dmin){
						dmin = nd2[qlist2[i]];
						dmins = qlist2[i];
					}
				}
			}
			if(dmins == -1){
				break;
			}
			ndf2[dmins] = 1;
			qlist2.Remove(dmins);
			for(var i=0;i<m;i++){
				if(a[i,0]-1==dmins){
					if(ndf2[a[i,1]-1] == 0){
						nd2[a[i,1]-1] = Math.Min(nd2[a[i,1]-1],dmin+a[i,3]);
						qlist2.Add(a[i,1]-1);
					}
				} else if(a[i,1]-1==dmins){
					if(ndf2[a[i,0]-1] == 0){
						nd2[a[i,0]-1] = Math.Min(nd2[a[i,0]-1],dmin+a[i,3]);
						qlist2.Add(a[i,0]-1);
					}
				}
			}
		}
		long[]ys = new long[n];
		ys[n-1] = nd[n-1]+nd2[n-1];
		for(var i=n-2;i>=0;i--){
			ys[i] = Math.Min(ys[i+1],nd[i]+nd2[i]);
		}
		
		var sw = new StreamWriter(Console.OpenStandardOutput()){AutoFlush = false};
		Console.SetOut(sw);
		
		for(var i=0;i<n;i++){
			Console.WriteLine(Math.Pow(10,15)-ys[i]);
			
		Console.Out.Flush();
		}
	}
}