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: Runtime Error)

import heapq

def dijkstra(ad_lst, n, s, t):
    costs = [float('inf') for i in range(n+1)]
    costs[s] = 0
    prevs = [-1 for _ in range(n+1)]
    Q = []
    heapq.heappush(Q, [costs[s], s])

    while(Q != []):
        c, u = heapq.heappop(Q)
        for v, c in ad_lst[u]:
            alt = costs[u] + c
            if alt < costs[v]:
                heapq.heappush(Q, [alt, v])
                costs[v] = alt

    return costs

n,m,s,t, = map(int, input().split())

route_y = {i: [] for i in range(n+1)}
route_s = {i: [] for i in range(n+1)}
for i in range(m):
    u, v, a, b = map(int, input().split())
    route_y[u].append([v,a])
    route_y[v].append([u,a])
    route_s[u].append([v,b])
    route_s[v].append([u,b])

cost_y = dijkstra(route_y, n, s, t)
cost_s = dijkstra(route_s, n, t, s)
cost_ys = [[y+s, i] for y, s in zip(cost_y, cost_s)]

cost_ys.sort()
yen = 10**15
k = 0
for i in range(n):
    while cost_ys[k][1]<=i: k+=1
    print(yen-cost_ys[k][0])
# inf = float('inf')
# result = [inf for _ in range(n+1)]
# min_cost = result[-1]
# for i in range(n, 0, -1):
#     if min_cost > cost_ys[i]:
#         min_cost = cost_ys[i]
#     result[i] = min_cost
# yen = 10**15
# for i in range(1, n+1):
#     print(yen - result[i])