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.Linq;//リストの使用
using System.Collections.Generic;
class Program
{
static long[] memo = new long[2];
static string[] input = Console.ReadLine().Split(' ');//Splitで区切り文字を指定して複数個受け取る。
static long n = long.Parse(input[0]);
static long m = long.Parse(input[1]);
static long s = long.Parse(input[2])-1;
static long t = long.Parse(input[3])-1;
static List<long>[] nextList = new List<long>[n];
static List<long[]> distanceQue = new List<long[]>();
static long[] distanceMemo = new long[n];
static long[] distanceS = new long[n];
static long[] distanceG = new long[n];

	static void Main()
	{
  for(int i = 0; i < n; i++)
    {
    nextList[i] = new List<long>();//リストの宣言
    distanceMemo[i] = 999999999999;
    }
  for(int i = 0; i < m; i++)
    {
    long[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);//１行ごとに受け取る
    nextList[nums[0]-1].Add(nums[1]-1);//隣接リストへ追加。
    nextList[nums[0]-1].Add(nums[2]);
    nextList[nums[0]-1].Add(nums[3]);
    nextList[nums[1]-1].Add(nums[0]-1);
    nextList[nums[1]-1].Add(nums[2]);
    nextList[nums[1]-1].Add(nums[3]);
    }

  distanceQue.Add(new long[2]{0,s});
  distanceMemo[s] = 0;
  
  while(distanceQue.Count() > 0)//Sからの最短距離
    {
    //Console.WriteLine(distanceMemo[0] + " " +distanceMemo[1] + " " +distanceMemo[2] + " " +distanceMemo[3]);
    Console.WriteLine(memo[0] + " " + memo[1]);
      PopHeap(distanceQue);
      distanceMemo[memo[1]] = memo[0];//memoに(距離、頂点番号)を保存
      if(memo[0] > distanceMemo[memo[1]]) continue;
      for(int j = 0; j < nextList[memo[1]].Count(); j+=3)
      {
        long a = (long)((List<long>)nextList[memo[1]])[j];//要素の取得。intだと65535までになる。 
        long b = (long)((List<long>)nextList[memo[1]])[j+1];//距離
        if(distanceMemo[a] > distanceMemo[memo[1]] + b)
        {
          distanceMemo[a] = distanceMemo[memo[1]] + b;
          PushHeap(distanceQue,new long[2]{distanceMemo[a], a});
        }
      }
    }
    
	Console.WriteLine(distanceMemo[0] + " " +distanceMemo[1] + " " +distanceMemo[2] + " " +distanceMemo[3]);
	}

  //引数は、昇順にソート済みのリスト
  static List<long[]> PushHeap(List<long[]> nums, long[] addNum)//現在のヒープnumsにaddNumを追加
  {
    int n = nums.Count;
    nums.Add(addNum);

    while (n > 0)
    {
      int i = (n-1)/2;
      if (nums[n][0] < nums[i][0])//親のほうが大きいなら入れ替え。
      {
        long tmp = nums[n][0];
        long tmpp = nums[n][1];
        nums[n][0] = nums[i][0];
        nums[n][1] = nums[i][1];
        nums[i][0] = tmp;
        nums[i][1] = tmpp;
      }
      n = i;
    }
    return nums;
  }

  static List<long[]> PopHeap(List<long[]> nums)//ヒープnumsからポップ
  {
    int n = nums.Count - 1;
    memo = nums[0];//捨てる値の保存
    nums[0][0] = nums[n][0];//最小値である根を削除してリストの最後を持ってくる
    nums[0][1] = nums[n][1];
    nums.RemoveAt(n);

    for (int i = 0, j; (j = 2*i+1) < n; )
    {
      if ((j != n-1) && (nums[j][0] > nums[j+1][0]))// ２つの子のうち、値の小さい方を選ぶ
        j++;
      if (nums[i][0] > nums[j][0])//子が親より小さいなら入れ替え
      {
        long tmp = nums[j][0];
        long tmpp = nums[j][1];
        nums[j][0] = nums[i][0];
        nums[j][1] = nums[i][1];
        nums[i][0] = tmp;
        nums[i][1] = tmpp;
      }
      i = j;
    }
    return nums;
  }
}