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 Python (3.4.3) (judge verdict: Accepted)

import heapq
INF=10**15
class weighted_graph():
    Vertex= 1
    AdjList=[dict() for i in range(Vertex)]
    Edge=[]
    directed=False
    def __init__(self,size,directed=False):
        self.directed=directed
        self.Vertex=size
        self.AdjList=[dict() for i in range(self.Vertex)]
    def add_edge(self,start,finish,weight):
        self.AdjList[start][finish]=weight
        self.Edge.append((start,finish,weight))
        if not(self.directed):
            self.AdjList[finish][start]=weight
    def Dijkstra(self, st):
        dist=[INF for i in range(N)]
        dist[st]=0
        q=[]
        heapq.heappush(q,(0,st))
        while (len(q)!=0):
            prc,src=heapq.heappop(q)
            if dist[src]<prc:
                continue
            for dest in self.AdjList[src]:
                cost=self.AdjList[src][dest]
                if dist[dest]>dist[src]+cost:
                    dist[dest]=dist[src]+cost
                    heapq.heappush(q,(dist[dest],dest))                
        return dist
    def Warshall_Floyd(self):
        d=[[INF for i in range(self.Vertex)] for i in range(self.Vertex)]
        for seq in self.Edge:
            d[seq[0]][seq[1]]=seq[2]
            if not(self.directed):
                d[seq[1]][seq[0]]=seq[2]
        for k in range(self.Vertex):
            for i in range(self.Vertex):
                for j in range(self.Vertex):
                    if d[i][j]>d[i][k]+d[k][j]:
                        d[i][j]=d[i][k]+d[k][j]
        return d
N,M,s,t=map(int,input().split())
s-=1;t-=1
G=weighted_graph(N,directed=False)
H=weighted_graph(N,directed=False)
for _ in range(M):
    u,v,a,b=map(int,input().split())
    G.add_edge(u-1,v-1,a)
    H.add_edge(u-1,v-1,b)
dist1=G.Dijkstra(s)
dist2=H.Dijkstra(t)
tmp=[INF-dist1[i]-dist2[i] for i in range(N)]
ans=[0 for i in range(N)]
ans[-1]=tmp[-1]
for i in range(N-1)[::-1]:
    ans[i]=max(ans[i+1],tmp[i])
for i in ans:
    print(i)