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.Linq;

namespace SoundHound2018D
{
    internal class Program
    {
        public static void Main(string[] args)
        {
            int[] inp = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            int n = inp[0];
            int m = inp[1];
            int s = inp[2] - 1;
            int t = inp[3] - 1;

            int[] u = new int[m];
            int[] v = new int[m];
            int[] a = new int[m];
            int[] b = new int[m];
            for (int i = 0; i < m; i++)
            {
                int[] train = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
                u[i] = train[0] - 1;
                v[i] = train[1] - 1;
                a[i] = train[2];
                b[i] = train[3];
            }

            long[] yen = new long[n];
            long[] snuke = new long[n];
            for (int i = 0; i < n; i++)
            {
                yen[i] = Int64.MaxValue;
                snuke[i] = Int64.MaxValue;
            }

            yen[s] = 0;
            snuke[t] = 0;

            Dictionary<long, List<long[]>> tree = new Dictionary<long, List<long[]>>();
            for (int i = 0; i < m; i++)
            {
                if (!tree.ContainsKey(u[i]))
                {
                    tree.Add(u[i], new List<long[]>());
                }

                if (!tree.ContainsKey(v[i]))
                {
                    tree.Add(v[i], new List<long[]>());
                }

                tree[u[i]].Add(new long[] {v[i], a[i], b[i]});
                tree[v[i]].Add(new long[] {u[i], a[i], b[i]});
            }

            Dictionary<long, List<long[]>> tree1 = new Dictionary<long, List<long[]>>(tree);
            Dictionary<long, List<long[]>> tree2 = new Dictionary<long, List<long[]>>(tree);
            List<long> queue = new List<long>();
            queue.Add(s);
            while (queue.Count > 0)
            {
                long current = queue.First();
                queue.Remove(current);

                if (tree1.ContainsKey(current))
                {
                    List<long[]> aaa = tree1[current];
                    tree1.Remove(current);
                    foreach (var temp in aaa)
                    {
                        queue.Add(temp[0]);
                        yen[temp[0]] = Math.Min(yen[temp[0]], yen[current] + temp[1]);
                    }

                }
            }

            queue.Add(t);

            while (queue.Count > 0)
            {
                long current = queue.First();
                queue.Remove(current);

                if (tree2.ContainsKey(current))
                {
                    List<long[]> aaa = tree2[current];
                    tree2.Remove(current);
                    foreach (var temp in aaa)
                    {
                        queue.Add(temp[0]);
                        snuke[temp[0]] = Math.Min(snuke[temp[0]], snuke[current] + temp[2]);
                    }
                }
            }
            long[] res = new long[n];
            long here = (long) Math.Pow(10, 15);
            for (int i = 0; i < n; i++)
            {
                res[i] = here - yen[i] - snuke[i];
            }

            for (int i = 0; i < n; i++)
            {
                Console.WriteLine(res.Max());
                res[i] = 0;
            }
        }
    }
}