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.Collections.Generic;
using System.Text;

struct Edge
{
    public int V, A, B;

    public Edge(int v, int a, int b)
    {
        V = v;
        A = a;
        B = b;
    }
}

#region PriorityQueue

/// <summary>
/// 指定した型のインスタンスを最も価値が低い順に取り出すことが可能な可変サイズのコレクションを表します．
/// </summary>
/// <typeparam name="T">優先度付きキュー内の要素の型を指定します．</typeparam>
/// <remarks>内部的にはバイナリヒープによって実装されています．</remarks>
public class PriorityQueue<T>
{
    readonly List<T> heap = new List<T>();
    readonly Comparison<T> cmp;

    /// <summary>
    /// デフォルトの比較子を使用してインスタンスを初期化します．
    /// </summary>
    /// <remarks>この操作は O(1) で実行されます．</remarks>
    public PriorityQueue()
    {
        cmp = Comparer<T>.Default.Compare;
    }

    /// <summary>
    /// デリゲートで表されるような比較関数を使用してインスタンスを初期化します．
    /// </summary>
    /// <param name="comparison"></param>
    /// <remarks>この操作は O(1) で実行されます．</remarks>
    public PriorityQueue(Comparison<T> comparison)
    {
        cmp = comparison;
    }

    /// <summary>
    /// 指定された比較子を使用してインスタンスを初期化します．
    /// </summary>
    /// <param name="comparer"></param>
    /// <remarks>この操作は O(1) で実行されます．</remarks>
    public PriorityQueue(IComparer<T> comparer)
    {
        cmp = comparer.Compare;
    }

    /// <summary>
    /// 優先度付きキューに要素を追加します．
    /// </summary>
    /// <param name="item">優先度付きキューに追加される要素</param>
    /// <remarks>最悪計算量 O(log N) で実行されます．</remarks>
    public void Enqueue(T item)
    {
        var pos = heap.Count;
        heap.Add(item);
        while (pos > 0)
        {
            var par = (pos - 1) / 2;
            if (cmp(heap[par], item) <= 0)
                break;
            heap[pos] = heap[par];
            pos = par;
        }

        heap[pos] = item;
    }

    /// <summary>
    /// 優先度付きキューから最も価値が低い要素を削除し，返します．
    /// </summary>
    /// <returns>優先度付きキューから削除された要素．</returns>
    /// <remarks>最悪計算量 O(log N) で実行されます．</remarks>
    public T Dequeue()
    {
        var ret = heap[0];
        var pos = 0;
        var x = heap[heap.Count - 1];

        while (pos * 2 + 1 < heap.Count - 1)
        {
            var lch = pos * 2 + 1;
            var rch = pos * 2 + 2;
            if (rch < heap.Count - 1 && cmp(heap[rch], heap[lch]) < 0) lch = rch;
            if (cmp(heap[lch], x) >= 0)
                break;
            heap[pos] = heap[lch];
            pos = lch;
        }

        heap[pos] = x;
        heap.RemoveAt(heap.Count - 1);
        return ret;
    }

    /// <summary>
    ///  優先度付きキューに含まれる最も価値が低い要素を削除せずに返します．
    /// </summary>
    /// <returns>優先度付きキューに含まれる最も価値が低い要素．</returns>
    /// <remarks>この操作は O(1) で実行されます．</remarks>
    public T Peek()
    {
        return heap[0];
    }

    /// <summary>
    /// 優先度付きキュー内の要素の数を取得します．
    /// </summary>
    /// <returns>優先度付キュー内にある要素の数</returns>
    /// <remarks>最悪計算量 O(1) で実行されます．</remarks>
    public int Count
    {
        get { return heap.Count; }
    }

    /// <summary>
    /// 優先度付きキュー内に要素が存在するかどうかを O(1) で判定します．
    /// </summary>
    /// <returns>優先度付キュー内にある要素が存在するならば true，そうでなければ　false．</returns>
    /// <remarks>この操作は O(1) で実行されます．</remarks>
    public bool Any()
    {
        return heap.Count > 0;
    }

    /// <summary>
    /// 優先度付きキューに含まれる要素を昇順に並べて返します．
    /// </summary>
    /// <remarks>この操作は計算量 O(N log N)で実行されます．</remarks>
    public T[] Items
    {
        get
        {
            var ret = heap.ToArray();
            Array.Sort(ret, cmp);
            return ret;
        }
    }
}

#endregion

struct P
{
    public int N;
    public long Dist;

    public P(int n, long d)
    {
        N = n;
        Dist = d;
    }
}

public class Program
{
    private int N, M, S, T;
    private List<Edge>[] E;

    // Sから各駅までの円での最小コスト
    private long[] YenCost;

    // Tから各駅までスヌークでの最小コスト
    private long[] SnuukCost;

    private void Scan()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        M = sc.NextInt();
        S = sc.NextInt() - 1;
        T = sc.NextInt() - 1;
        E = new List<Edge>[N];

        for (int i = 0; i < N; i++)
        {
            E[i] = new List<Edge>();
        }

        for (int i = 0; i < M; i++)
        {
            int u = sc.NextInt() - 1;
            int v = sc.NextInt() - 1;
            int a = sc.NextInt();
            int b = sc.NextInt();
            E[u].Add(new Edge(v, a, b));
            E[v].Add(new Edge(u, a, b));
        }
    }

    private void CalcYenCost()
    {
        YenCost = new long[N];
        for (int i = 0; i < N; i++)
        {
            YenCost[i] = long.MaxValue;
        }

        var pq = new PriorityQueue<P>((a, b) => a.Dist.CompareTo(b.Dist));
        pq.Enqueue(new P(S, 0));
        YenCost[S] = 0;

        for (int i = 0; i < N; i++)
        {
            P d;
            while (true)
            {
                d = pq.Dequeue();
                if (YenCost[d.N] == d.Dist) break;
            }
            foreach (Edge edge in E[d.N])
            {
                int nextN = edge.V;
                long nextDist = d.Dist + edge.A;
                if (YenCost[nextN] < nextDist) continue;
                YenCost[nextN] = nextDist;
                pq.Enqueue(new P(nextN, nextDist));
            }
        }
    }

    private void CalcSnuukCost()
    {
        SnuukCost = new long[N];
        for (int i = 0; i < N; i++)
        {
            SnuukCost[i] = long.MaxValue;
        }

        var pq = new PriorityQueue<P>((a, b) => a.Dist.CompareTo(b.Dist));
        pq.Enqueue(new P(T, 0));
        SnuukCost[T] = 0;
        for(int i = 0;i<N;i++)
        {
            P d;
            while (true)
            {
                d = pq.Dequeue();
                if (SnuukCost[d.N] == d.Dist) break;
            }
            foreach (Edge edge in E[d.N])
            {
                int nextN = edge.V;
                long nextDist = d.Dist + edge.B;
                if (SnuukCost[nextN] < nextDist) continue;
                SnuukCost[nextN] = nextDist;
                pq.Enqueue(new P(nextN, nextDist));
            }
        }
    }

    private void Write()
    {
        long min = long.MaxValue;
        long[] ans = new long[N];
        for (int i = N - 1; i >= 0; i--)
        {
            // cost[i] 駅i+1で両替ができるので 今までの最小コストか s->i+1(yen)まで + i+1->t(snuuk)まで　の小さい方
            min = Math.Min(min, YenCost[i] + SnuukCost[i]);
            ans[i] = 1000000000000000 - min;
        }

        Console.WriteLine(string.Join("\n", ans));
    }


    public void Solve()
    {
        Scan();
        CalcYenCost();
        CalcSnuukCost();
        Write();
    }


    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}