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 (3.8.2) (judge verdict: Accepted)

import heapq
INF = float('inf')


def dijkstra(s, n, edges, d):
    d = [INF] * n
    d[s] = 0
    pq = [(0, s)]
    while len(pq):
        c, v = heapq.heappop(pq)
        if d[v] < c:
            continue
        for v2nvcost, nv in edges[v]:
            if c + v2nvcost < d[nv]:
                d[nv] = c + v2nvcost
                heapq.heappush(pq, (d[nv], nv))
    return d


n, m, s, t = map(int, input().split())
s -= 1
t -= 1
edges_yen = [[] for _ in range(n)]
edges_snuuk = [[] for _ in range(n)]
for _ in range(m):
    u, v, a, b = map(int, input().split())
    u -= 1
    v -= 1
    edges_yen[u].append((a, v))
    edges_yen[v].append((a, u))
    edges_snuuk[u].append((b, v))
    edges_snuuk[v].append((b, u))
d = [INF] * n
d_yen = dijkstra(s, n, edges_yen, d)
d = [INF] * n
d_snuuk = dijkstra(t, n, edges_snuuk, d)
temp = [0] * n  # 場所iで両替した時の最終スヌーク
for i in range(n):
    temp[i] = 10**15 - (d_yen[i] + d_snuuk[i])

ans = [INF] * n
ans[-1] = temp[-1]
for i in range(n - 2, -1, -1):
    ans[i] = max(ans[i + 1], temp[i])
# print(d_yen)
# print(d_snuuk)
for i in range(n):
    print(ans[i])