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;

public class PriorityQueue<TKey, TValue>
{
    SortedDictionary<TKey, List<TValue>> dict = new SortedDictionary<TKey, List<TValue>>();
    Dictionary<TValue, TKey> rDict = new Dictionary<TValue, TKey>();
    int count = 0;

    public int Count
    {
        get
        {
            return count;
        }
    }

    public void Enqueue(TKey key, TValue value)
    {
        List<TValue> set;
        if (!dict.TryGetValue(key, out set))
        {
            set = new List<TValue>();
            dict[key] = set;
        }

        set.Add(value);
        rDict[value] = key;
        count++;
    }

    public KeyValuePair<TKey, TValue> Dequeue()
    {
        var queue = dict.First();
        if (queue.Value.Count <= 1)
        {
            dict.Remove(queue.Key);
        }

        var val = queue.Value.First();
        queue.Value.Remove(val);
        count--;
        rDict.Remove(val);
        return new KeyValuePair<TKey, TValue>(queue.Key, val);
    }

    public KeyValuePair<TKey, TValue> Peek()
    {
        var queue = dict.First();
        var val = queue.Value.First();
        return new KeyValuePair<TKey, TValue>(queue.Key, val);
    }

    public void ChangePriority(TKey newKey, TValue value)
    {
        ChangePriority(rDict[value], newKey, value);
    }

    public void ChangePriority(TKey prevKey, TKey newKey, TValue value)
    {
        var set = dict[prevKey];
        set.Remove(value);
        if (set.Count == 0)
        {
            dict.Remove(prevKey);
        }
        count--;

        Enqueue(newKey, value);
    }

    public bool ContainsValue(TValue val)
    {
        return rDict.ContainsKey(val);
    }
}

public class Node
{
    public long Yen;
    public long Snuuk;
    public List<Train> Trains = new List<Train>();
}

public class Train
{
    public int Dest;
    public long Yen;
    public long Snuuk;

    public Train(int dest, int yen, int snuuk)
    {
        Dest = dest;
        Yen = yen;
        Snuuk = snuuk;
    }
}

class Solution
{
    public static long[] Dijkstra(IList<Tuple<int, long>>[] link, int start)
    {
        var n = link.Length;
        var d = Enumerable.Repeat(long.MaxValue, n).ToArray();
        d[start] = 0;
        var q = new PriorityQueue<long, int>();
        q.Enqueue(0, start);

        while (q.Count > 0)
        {
            var u = q.Dequeue().Value;
            foreach (var nei in link[u])
            {
                var v = nei.Item1;
                var len = nei.Item2;
                if (d[v] > d[u] + len)
                {
                    d[v] = d[u] + len;
                    if (q.ContainsValue(v))
                    {
                        q.ChangePriority(d[v], v);
                    }
                    else
                    {
                        q.Enqueue(d[v], v);
                    }
                }
            }
        }

        return d;
    }

    const long MaxYen = 1000000000000000;

    static void Main()
    {
        var vals = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
        int n = vals[0];
        int m = vals[1];
        int start = vals[2];
        int goal = vals[3];

        var nodes = Enumerable.Range(0, n + 1).Select(_ => new Node()).ToArray();
        for (int i = 0; i < m; i++)
        {
            vals = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
            var u = vals[0];
            var v = vals[1];
            var yen = vals[2];
            var snuuk = vals[3];
            nodes[u].Trains.Add(new Train(v, yen, snuuk));
            nodes[v].Trains.Add(new Train(u, yen, snuuk));
        }

        var yens = Dijkstra(nodes.Select(node => node.Trains.Select(t => Tuple.Create(t.Dest, t.Yen)).ToList()).ToArray(), start);
        var snuuks = Dijkstra(nodes.Select(node => node.Trains.Select(t => Tuple.Create(t.Dest, t.Snuuk)).ToList()).ToArray(), goal);
        for (int i = 0; i < nodes.Length; i++)
        {
            var node = nodes[i];
            node.Yen = yens[i];
            node.Snuuk = snuuks[i];
        }

        var resultQueue = new Queue<Tuple<long, int>>(nodes.Select((node, i) => Tuple.Create(MaxYen - node.Yen - node.Snuuk, i)).Skip(1).OrderByDescending(item => item.Item1));

        var answer = new StringBuilder();
        for (int year = 0; year < n; year++)
        {
            while (resultQueue.Peek().Item2 <= year)
            {
                resultQueue.Dequeue();
            }

            var best = resultQueue.Peek();
            answer.AppendLine(best.Item1.ToString());
        }

        Console.WriteLine(answer);
    }
}