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: Runtime Error)

using System;
using System.Collections.Generic;
class Program
{
	static void Main(string[] args)
	{
		string[] str = Console.ReadLine().Split();
		long n = long.Parse(str[0]);
		long m = long.Parse(str[1]);
		long s = long.Parse(str[2]);
		long t = long.Parse(str[3]);
		List<long>[] ma = new List<long>[n];
		for(var i=0;i<n;i++){
			ma[i] = new List<long>();
		}
		long[,]a = new long[m,4];
		for(var i=0;i<m;i++){
			string[] str2 = Console.ReadLine().Split();
			a[i,0] = long.Parse(str2[0]);
			a[i,1] = long.Parse(str2[1]);
			a[i,2] = long.Parse(str2[2]);
			a[i,3] = long.Parse(str2[3]);
			ma[a[i,0]-1].Add(i);
			ma[a[i,1]-1].Add(i);
		}
		var dq = new Dijkstra(n);
		for(var i=0;i<m;i++){
			dq.AddPath(a[i,0],a[i,1],a[i,2]);
		}
		var dq2 = new Dijkstra(n);
		for(var i=0;i<m;i++){
			dq2.AddPath(a[i,0],a[i,1],a[i,3]);
		}
		long[]ys = new long[n];
		ys[n-1] = dq.GetMinCost(s, n)+dq2.GetMinCost(t, n);
		for(var i=n-2;i>=0;i--){
			ys[i] = Math.Min(ys[i+1],dq.GetMinCost(s, i+1)+dq2.GetMinCost(t, i+1));
		}
		for(var i=0;i<n;i++){
			Console.WriteLine(Math.Pow(10,15)-ys[i]);
		}
	}
}
 
	public class PriorityQueue<T> where T: IComparable{
        private IComparer<T> _comparer = null;
        private long _type = 0;
        private T[] _heap;
        private long _sz = 0;
        private long _count = 0;
        /// Priority Queue with custom comparer
        public PriorityQueue(long maxSize, IComparer<T> comparer){
            _heap = new T[maxSize];
            _comparer = comparer;
        }
        /// Priority queue
        /// <param name="maxSize">max size</param>
        /// <param name="type">0: asc, 1:desc</param>
        public PriorityQueue(long maxSize, long type = 0){
            _heap = new T[maxSize];
            _type = type;
        }
        private long Compare(T x, T y){
            if (_comparer != null) return _comparer.Compare(x, y);
            return _type == 0 ? x.CompareTo(y) : y.CompareTo(x);
        }
        public void Push(T x){
            _count++;
            //node number
            var i = _sz++;
            while (i > 0){
                //parent node number
                var p = (i - 1) / 2;
                if (Compare(_heap[p], x) <= 0) break;
                _heap[i] = _heap[p];
                i = p;
            }
            _heap[i] = x;
        }
        public T Pop(){
            _count--;
            T ret = _heap[0];
            T x = _heap[--_sz];
            long i = 0;
            while (i * 2 + 1 < _sz){
                //children
                long a = i * 2 + 1;
                long b = i * 2 + 2;
                if (b < _sz && Compare(_heap[b], _heap[a]) < 0) a = b;
                if (Compare(_heap[a], x) >= 0) break;
                _heap[i] = _heap[a];
                i = a;
            }
            _heap[i] = x;
            return ret;
        }
        public long Count(){
            return _count;
        }
        public T Peek(){
            return _heap[0];
        }
        public bool Contains(T x){
            for (long i = 0; i < _sz; i++) if (x.Equals(_heap[i])) return true;
            return false;
        }
        public void Clear(){
            while (this.Count() > 0) this.Pop();
        }
        public IEnumerator<T> GetEnumerator(){
            var ret = new List<T>();
            while (this.Count() > 0){
                ret.Add(this.Pop());
            }
            foreach (var r in ret){
                this.Push(r);
                yield return r;
            }
        }
        public T[] ToArray(){
            T[] array = new T[_sz];
            long i = 0;
            foreach (var r in this){
                array[i++] = r;
            }
            return array;
        }
    }

    /// Get min cost between two points
    public class Dijkstra{
        private long maxIndex = -1;
        private const long INF = Int64.MaxValue;
        private long[,] _edgeArray;
        /// Dijkstra, get min cost between two points
        /// should not contain negatie cost path
        /// <param name="size">max index of vertices</param>
        public Dijkstra(long size){
            maxIndex = size + 1;
            _edgeArray = new long[maxIndex, maxIndex];
            for (long i = 0; i < _edgeArray.GetLength(0); i++){
                for (long j = 0; j < _edgeArray.GetLength(1); j++){
                    _edgeArray[i, j] = INF;
                    if (i == j) _edgeArray[i, j] = 0;
                }
            }
        }
        // Add a path(no direction) with its cost
        public void AddPath(long s, long t, long cost){
            _edgeArray[s, t] = Math.Min(_edgeArray[s, t], cost);
            _edgeArray[t, s] = _edgeArray[s, t];
        }
        //Get the min cost between s and t
        //return Int32.MaxValue if no path
        public long GetMinCost(long s, long t){
            long[] cost = new long[maxIndex];
            for (long i = 0; i < cost.Length; i++) cost[i] = INF;
            cost[s] = 0;
            var priorityQueue = new PriorityQueue<ComparablePair<long, long>>(maxIndex);
            priorityQueue.Push( new ComparablePair<long, long>(0, s) );
            while (priorityQueue.Count() > 0){
                var costDestinationPair = priorityQueue.Pop();
                if (cost[costDestinationPair.Item2] < costDestinationPair.Item1) continue;
                for (long i = 0; i < maxIndex; i++){
                    long newCostToi = _edgeArray[costDestinationPair.Item2, i] == INF ? INF :
                        costDestinationPair.Item1 + _edgeArray[costDestinationPair.Item2, i];
                    if (newCostToi < cost[i]){
                        cost[i] = newCostToi;
                        priorityQueue.Push(new ComparablePair<long, long>(newCostToi, i));
                    }
                }
            }
            return cost[t];
        }
    }
    public class ComparablePair<T, U> : IComparable where T : IComparable<T>{
        public readonly T Item1;
        public readonly U Item2;
        public int CompareTo(object obj){
            ComparablePair<T, U> castedObj = (ComparablePair<T, U>)obj;
            return this.Item1.CompareTo(castedObj.Item1);
        }
        public ComparablePair(T t, U u){
            Item1 = t;
            Item2 = u;
        }
    }