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 PyPy3 (2.4.0) (judge verdict: Accepted)

import heapq
import sys

input = sys.stdin.readline
INF = 10**18

n,m,s,t = map(int,input().split())
edge = [list(map(int,input().split())) for _ in range(m)]

graph_s = [[] for _ in range(n)]
graph_t = [[] for _ in range(n)]

for e in edge:
    graph_s[e[0]-1].append([e[1]-1,e[2]])
    graph_s[e[1]-1].append([e[0]-1,e[2]])
    graph_t[e[0]-1].append([e[1]-1,e[3]])
    graph_t[e[1]-1].append([e[0]-1,e[3]])

dist_s = [INF]*n
dist_t = [INF]*n

dist_s[s-1] = 0
dist_t[t-1] = 0

heap_s = [[0,s-1]]
heapq.heapify(heap_s)

heap_t = [[0,t-1]]
heapq.heapify(heap_t)

while heap_s:
    cost,node = heapq.heappop(heap_s)
    if dist_s[node] < cost:
        continue
    for adj,adjcost in graph_s[node]:
        if dist_s[node] + adjcost < dist_s[adj]:
            dist_s[adj] = dist_s[node] + adjcost
            heapq.heappush(heap_s,[dist_s[adj],adj])

while heap_t:
    cost,node = heapq.heappop(heap_t)
    if dist_t[node] < cost:
        continue
    for adj,adjcost in graph_t[node]:
        if dist_t[node] + adjcost < dist_t[adj]:
            dist_t[adj] = dist_t[node] + adjcost
            heapq.heappush(heap_t,[dist_t[adj],adj])

dist = INF
ans = []
                                              
for i in range(n):
    dist = min(dist,dist_t[-1-i]+dist_s[-1-i])
    ans.append(10**15-dist)
    
for i in range(n):
    print(ans[-1-i])