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.Linq;
using System.Collections.Generic;

using E = System.Linq.Enumerable;

public class Program
{
    public static void Main()
    {
        var nmst = Console.ReadLine().Split().Select(x => int.Parse(x)).ToArray();
        var n = nmst[0];
        var m = nmst[1];
        var s = nmst[2] - 1;
        var t = nmst[3] - 1;
        var yen = E.Range(0, n).Select(_ => new Dictionary<int, long>()).ToArray();
        var snk = E.Range(0, n).Select(_ => new Dictionary<int, long>()).ToArray();
        for (int i = 0; i < m; i++)
        {
            var uvab = Console.ReadLine().Split().Select(x => int.Parse(x)).ToArray();
            if (uvab[0] == uvab[1])
            {
                continue;
            }
            uvab[0] -= 1;
            uvab[1] -= 1;
            yen[uvab[0]][uvab[1]] = yen[uvab[1]][uvab[0]] = uvab[2];
            snk[uvab[0]][uvab[1]] = snk[uvab[1]][uvab[0]] = uvab[3];
        }

        var min = E.Zip(Dijkstra(n, yen, s), Dijkstra(n, snk, t), (x, y) => x + y)
            .Select((d, i) => new { d, i })
            .OrderBy(x => x.d)
            .ToArray();
        for (int i = 0; i < n; i++)
        {
            Console.WriteLine((long)1e15 - min.First(x => x.i >= i).d);
        }
    }

    static long[] Dijkstra(int n, Dictionary<int, long>[] edges, int s)
    {
        var pryq = new SortedDictionary<long, Queue<int>>() { { long.MaxValue, new Queue<int>(E.Range(0, n).Where(x => x != s)) } };
        pryq[0] = new Queue<int>(new [] { s });
        var d = new long[n];
        for (int i = 0; i < n; i++)
        {
            d[i] = i == s ? 0 : long.MaxValue;
        }

        while (pryq.Any())
        {
            var q = pryq.First();
            var u = q.Value.Dequeue();
            if (!q.Value.Any())
            {
                pryq.Remove(q.Key);
            }

            foreach (var e in edges[u])
            {
                var x = d[u] + e.Value;
                if (d[e.Key] > x)
                {
                    d[e.Key] = x;
                    Queue<int> vs;
                    if (pryq.TryGetValue(x, out vs))
                    {
                        vs.Enqueue(e.Key);
                    }
                    else
                    {
                        pryq[x] = new Queue<int>(new [] { e.Key });
                    }
                }
            }
        }

        return d;
    }
}