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: Time Limit Exceeded)

n, m, s, t = map(int, input().split())
s, t = s-1, t-1
route_yen = [[0 for i in range(n)] for j in range(n)]
route_snook = [[0 for i in range(n)] for j in range(n)]
for i in range(m):
    ui, vi, ai, bi = map(int, input().split())
    ui, vi = vi-1, ui-1
    route_yen[ui][vi], route_yen[vi][ui] = ai, ai
    route_snook[ui][vi], route_snook[vi][ui] = bi, bi

node_num = n #ノードの数

unsearched_nodes = list(range(node_num)) # 未探索ノード
distance_yen = [10**15] * node_num # ノードごとの距離のリスト
previous_nodes = [-1] * node_num # 最短経路でそのノードのひとつ前に到達するノードのリスト
distance_yen[s] = 0 # 初期のノードの距離は0とする

# @GDaigo さんのコメントより関数の追加 2017/03/18
def get_target_min_index(min_index, distance, unsearched_nodes):
    start = 0
    while True:
        index = distance.index(min_index, start)
        found = index in unsearched_nodes
        if found:
            return index
        else:
            start = index + 1

while(len(unsearched_nodes) != 0): #未探索ノードがなくなるまで繰り返す
    # まず未探索ノードのうちdistanceが最小のものを選択する
    posible_min_distance = 10**15 # 最小のdistanceを見つけるための一時的なdistance。初期値は inf に設定。
    for node_index in unsearched_nodes: # 未探索のノードのループ
        if posible_min_distance > distance_yen[node_index]: 
            posible_min_distance = distance_yen[node_index] # より小さい値が見つかれば更新
    target_min_index = get_target_min_index(posible_min_distance, distance_yen, unsearched_nodes) # 未探索ノードのうちで最小のindex番号を取得
    unsearched_nodes.remove(target_min_index) # ここで探索するので未探索リストから除去

    target_edge = route_yen[target_min_index] # ターゲットになるノードからのびるエッジのリスト
    for index, route_dis in enumerate(target_edge):
        if route_dis != 0:
            if distance_yen[index] > (distance_yen[ target_min_index] + route_dis):
                distance_yen[index] = distance_yen[ target_min_index] + route_dis # 過去に設定されたdistanceよりも小さい場合はdistanceを更新
                previous_nodes[index] =  target_min_index #　ひとつ前に到達するノードのリストも更新


unsearched_nodes = list(range(node_num)) # 未探索ノード
distance_snook = [10**15] * node_num # ノードごとの距離のリスト
previous_nodes = [-1] * node_num # 最短経路でそのノードのひとつ前に到達するノードのリスト
distance_snook[t] = 0 # 初期のノードの距離は0とする

while(len(unsearched_nodes) != 0): #未探索ノードがなくなるまで繰り返す
    # まず未探索ノードのうちdistanceが最小のものを選択する
    posible_min_distance = 10**15 # 最小のdistanceを見つけるための一時的なdistance。初期値は inf に設定。
    for node_index in unsearched_nodes: # 未探索のノードのループ
        if posible_min_distance > distance_snook[node_index]: 
            posible_min_distance = distance_snook[node_index] # より小さい値が見つかれば更新
    target_min_index = get_target_min_index(posible_min_distance, distance_snook, unsearched_nodes) # 未探索ノードのうちで最小のindex番号を取得
    unsearched_nodes.remove(target_min_index) # ここで探索するので未探索リストから除去

    target_edge = route_snook[target_min_index] # ターゲットになるノードからのびるエッジのリスト
    for index, route_dis in enumerate(target_edge):
        if route_dis != 0:
            if distance_snook[index] > (distance_snook[ target_min_index] + route_dis):
                distance_snook[index] = distance_snook[ target_min_index] + route_dis # 過去に設定されたdistanceよりも小さい場合はdistanceを更新
                previous_nodes[index] =  target_min_index #　ひとつ前に到達するノードのリストも更新


ans_lst = [0 for i in range(n)]
temp_min = 10**15
for i in range(n)[::-1]:
    now = distance_yen[i] + distance_snook[i]
    temp_min = min(temp_min, now)
    ans_lst[i] = 10**15 - temp_min

for i in range(n):
    print(ans_lst[i])