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.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 List<KeyValuePair<int, long>>()).ToArray();
        var snk = E.Range(0, n).Select(_ => new List<KeyValuePair<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]].Add(new KeyValuePair<int, long>(uvab[1], uvab[2]));
            yen[uvab[1]].Add(new KeyValuePair<int, long>(uvab[0], uvab[2]));
            snk[uvab[0]].Add(new KeyValuePair<int, long>(uvab[1], uvab[3]));
            snk[uvab[1]].Add(new KeyValuePair<int, long>(uvab[0], uvab[3]));
        }

        var paths = E.Zip(Dijkstra(n, yen, s), Dijkstra(n, snk, t), (x, y) => x + y).ToArray();
        var min = new long[n];
        min[n - 1] = paths[n - 1];
        for (int i = n - 1; i > 0; i--)
        {
            min[i - 1] = Math.Min(min[i], paths[i - 1]);
        }
        for (int i = 0; i < n; i++)
        {
            Console.WriteLine((long)1e15 - min[i]);
        }
    }

    static long[] Dijkstra(int n, List<KeyValuePair<int, long>>[] edges, int s)
    {
        var pryq = new FibonacciHeap<long, int>();
        for (int i = 0; i < s; i++)
        {
            pryq.Insert(long.MaxValue, i);
        }
        for (int i = s + 1; i < n; i++)
        {
            pryq.Insert(long.MaxValue, i);
        }
        pryq.Insert(0, s);

        var d = new long[n];
        for (int i = 0; i < n; i++)
        {
            d[i] = long.MaxValue;
        }
        d[s] = 0;

        while (pryq.Size > 0)
        {
            var u = pryq.ExtractMinimum()?.Value ?? -1;
            foreach (var e in edges[u])
            {
                var x = d[u] + e.Value;
                if (d[e.Key] > x)
                {
                    d[e.Key] = x;
                    pryq.Insert(x, e.Key);
                }
            }
        }

        return d;
    }
}

class FibonacciHeap<TKey, TValue>
{
    private LinkedList<Carrier> _trees = new LinkedList<Carrier>();
    private LinkedListNode<Carrier> _peekNode = null;

    public int Size { get; private set; } = 0;

    public FibonacciHeap()
    {
    }

    public void Insert(TKey key, TValue value)
    {
        this.Size++;

        var x = new Carrier(key, value);
        if (this._peekNode == null)
        {
            this._peekNode = this._trees.AddFirst(x);
        }
        else
        {
            var t = this._trees.AddBefore(this._peekNode, x);
            var ord = Comparer<TKey>.Default.Compare(this._peekNode.Value.Key, key);
            if (ord > 0)
            {
                this._peekNode = t;
            }
        }
    }

    public KeyValuePair<TKey, TValue>? ExtractMinimum()
    {
        if (this._peekNode == null)
        {
            return null;
        }

        var kvp = new KeyValuePair<TKey, TValue>(this._peekNode.Value.Key, this._peekNode.Value.Value);
        var point = this._peekNode.Next;

        Array.ForEach(this._peekNode.Value.Produce(), x => this._trees.AddBefore(this._peekNode, x));
        this._trees.Remove(this._peekNode);

        if (this._trees.Any())
        {
            this.Consolidate(point);
            this.FindMinimum();
        }

        this.Size--;
        return kvp;
    }

    private void Consolidate(LinkedListNode<Carrier> begin)
    {
        if (begin == null)
        {
            begin = this._trees.First;
        }

        var bucket = new LinkedListNode<Carrier>[(int)Math.Ceiling(Math.Log(this.Size, 2)) + 1];
        var current = begin;
        do
        {
            var next = current.Next ?? this._trees.First;

            for (int d = current.Value.Degree; bucket[d] != null; d++)
            {
                var child = bucket[d];
                bucket[d] = null;

                var ord = Comparer<TKey>.Default.Compare(child.Value.Key, current.Value.Key);
                if (ord < 0)
                {
                    var t = child;
                    child = current;
                    current = t;
                }
                if (child == begin)
                {
                    begin = child.Next ?? this._trees.First;
                }
                if (child == next)
                {
                    next = child.Next ?? this._trees.First;
                }
                this._trees.Remove(child);
                child.Value.JoinTo(current.Value);
            }
            bucket[current.Value.Degree] = current;

            current = next;
        } while (current != begin);
    }

    private void FindMinimum()
    {
        this._peekNode = this._trees.First;
        var current = this._trees.First?.Next;
        while (current != null)
        {
            var ord = Comparer<TKey>.Default.Compare(this._peekNode.Value.Key, current.Value.Key);
            if (ord > 0)
            {
                this._peekNode = current;
            }
            current = current.Next;
        }
    }

    sealed class Carrier
    {
        public TKey Key { get; }
        public TValue Value { get; }
        private List<Carrier> _children;

        public int Degree => this._children.Count;

        public Carrier(TKey key, TValue value)
        {
            this.Key = key;
            this.Value = value;
            this._children = new List<Carrier>();
        }

        public void JoinTo(Carrier parent) => parent._children.Add(this);
        public void Remove(Carrier child) => this._children.Remove(child);
        public Carrier[] Produce() => this._children.ToArray();
    }
}