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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class Program
{
  class Vertex
  {
    public int next;//隣接頂点
    public int distance;//隣接頂点との距離
    public int distanceR;
    public Vertex(int next, int distance, int distanceR)
    {
      this.next = next;
      this.distance = distance;
      this.distanceR = distanceR;
    }
  }
  
  static void Main()
  {
    int[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
    int n = nums[0];
    int m = nums[1];
    int s = nums[2] - 1;
    int t = nums[3] - 1;
    var vertexes = new List<Vertex>[n];
    for (int i = 0; i < n; i++) 
    { 
      vertexes[i] = new List<Vertex>(); 
    }
    for (int i = 0; i < m; i++)
    {
      int[] numsSub = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
      int u = numsSub[0] - 1;
      int v = numsSub[1] - 1;
      int a = numsSub[2];
      int b = numsSub[3];
      vertexes[u].Add(new Vertex(v, a, b));
      vertexes[v].Add(new Vertex(u, a, b));
    }
    foreach (long answer in Solve(n, m, s, t, vertexes))
    {
      Console.WriteLine(answer);
    }
  }

  static IEnumerable<long> Solve(int n, int m, int s, int t, List<Vertex>[] vertexes)
  {
    long[] distanceMemo = Enumerable.Repeat(long.MaxValue, n).ToArray();//9999999円代入
    {
      var que = new Queue<int>();
      que.Enqueue(s);
      distanceMemo[s] = 0;
      while (que.Count > 0)
      {
        int u = que.Dequeue();//queから取り出し
        foreach (var vertex in vertexes[u])
        {
          int v = vertex.next;
          if (distanceMemo[u] + vertex.distance < distanceMemo[v])
          {
            distanceMemo[v] = distanceMemo[u] + vertex.distance;
            que.Enqueue(v);//queに入れる
          }
        }
      }
    }
    long[] distanceMemoR = Enumerable.Repeat(long.MaxValue, n).ToArray();//ゴールからの長さに9999999代入
    {
      var que = new Queue<int>();
      que.Enqueue(t);
      distanceMemoR[t] = 0;
      while (que.Count > 0)
      {
        int u = que.Dequeue();
        foreach (var vertex in vertexes[u])
        {
          int v = vertex.next;
          if (distanceMemoR[u] + vertex.distanceR < distanceMemoR[v])
          {
            distanceMemoR[v] = distanceMemoR[u] + vertex.distanceR;
            que.Enqueue(v);
          }
        }
      }
    }
    var distanceTotal = new long[n];
    distanceTotal[n-1] = distanceMemo[n-1] + distanceMemoR[n-1];
    for (int i = n-2; i >= 0; i--)
    {
      distanceTotal[i] = Math.Min(distanceMemo[i] + distanceMemoR[i], distanceTotal[i+1]);
    }
    return distanceTotal.Select(a => 1000000000000000L - a);
  }
  
}