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)

from heapq import heapify,heappush,heappop

n,m,s,t = map(int,input().split())#順に都市の数、電車の数、スタート、ゴール
adj_list = [[]for i in range(0,(n+1),1,)]#隣接リスト(都市1~nをそのまま使いたいのでn+1の大きさで宣言して0は使わない)
for i in range(0,m,1):
    u,v,a,b = map(int,input().split())
    adj_list[u].append([v,a,b])#[0]行先[1]円[2]スヌーク
    adj_list[v].append([u,a,b])

D_INF = 10**15
d = [D_INF]*(n+1)#最速到達距離。でかい数で初期化
d[s] = 0#スタート
que = []
heapify(que)
heappush(que,(d[s],s))

while(len(que)):
    time,place = heappop(que)
    if time > d[place]:
        continue
    for i in range(0,len(adj_list[place]),1):
        if d[adj_list[place][i][0]] > time + adj_list[place][i][1]:#とりあえず円で
            d[adj_list[place][i][0]] = time + adj_list[place][i][1]
            heappush(que,(d[adj_list[place][i][0]],adj_list[place][i][0]))


d_sn = [D_INF]*(n+1)#最速到達距離。でかい数で初期化
d_sn[t] = 0#スタート
heappush(que,(d_sn[t],t))

while(len(que)):
    time,place = heappop(que)
    if time > d_sn[place]:
        continue
    for i in range(0,len(adj_list[place]),1):
        if d_sn[adj_list[place][i][0]] > time + adj_list[place][i][2]:
             d_sn[adj_list[place][i][0]] = time + adj_list[place][i][2]
             heappush(que,(d_sn[adj_list[place][i][0]],adj_list[place][i][0]))

for i in range(1,n+1,1):
    d[i]=d[i]+d_sn[i]
    heappush(que,(d[i],i))
minimum,minimum_place = heappop(que)

for i in range(0,n,1):
    if minimum_place <= i:
        while(len(que)):
            minimum,minimum_place = heappop(que)
            if minimum_place > i:
                break
    print(10**15-minimum)