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


def dijkstra(start: int, graph: list) -> list:
    """dijkstra法: 始点startから各頂点への最短距離を求める
    計算量: O((E+V)logV)
    """
    INF = 10 ** 18
    n = len(graph)
    dist = [INF] * n
    dist[start] = 0
    q = [(0, start)] # q = [(startからの距離, 現在地)]
    while q:
        d, v = heapq.heappop(q)
        if dist[v] < d:
            continue
        for nxt_v, cost in graph[v]:
            if dist[v] + cost < dist[nxt_v]:
                dist[nxt_v] = dist[v] + cost
                heapq.heappush(q, (dist[nxt_v], nxt_v))
    return dist


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

graph_y = [[] for i in range(n)]
graph_s = [[] for i in range(n)]
for u, v, a, b in info:
    u -= 1
    v -= 1
    graph_y[u].append((v, a))
    graph_y[v].append((u, a))
    graph_s[u].append((v, b))
    graph_s[v].append((u, b))
    
s -= 1
t -= 1
dist_y = dijkstra(s, graph_y)
dist_s = dijkstra(t, graph_s)

ans = 10 ** 18
ret = [10 ** 15] * n
for i in range(n)[::-1]:
    tmp = dist_y[i] + dist_s[i]
    ans = min(ans, tmp)
    ret[i] -= ans
for i in ret:
    print(i)