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)

import sys
fin = sys.stdin.readline
from collections import defaultdict
from heapq import heappop, heappush


class Vertex(object):
    def __init__(self, name, distance):
        self.name = name
        self.distance = distance


    def __lt__(self, other_vertex):
        return self.distance < other_vertex.distance
    

    def __le__(self, other_vertex):
        return self.distance <= other_vertex.distance
    

    def __gt__(self, other_vertex):
        return self.distance > other_vertex.distance
    

    def __ge__(self, other_vertex):
        return self.distance >= other_vertex.distance


def initialize(adj, s):
    MAX_NUM = float('inf')
    parent = {s: None}
    d = {vertex: MAX_NUM for vertex in adj.keys()}
    d[s] = 0
    return d, parent


# Dijkstra can work using a minheap which does not have decrease_key.
# and also we do not have to implement Vertex class.
# Instead, we can use (distance, vertex).
# in this case, the time it takes is O(ElgE).
# E is at most O(V^2), thus O(ElgE) = O(ElgV).
def dijkstra_simple(adj, w, s):
    d, parent = initialize(adj, s)
    priority_queue = []
    heappush(priority_queue, (d[s], s))
    while priority_queue:
        distance, source = heappop(priority_queue) 
        for neighbor in adj[source]:
            new_distance = distance + w[(source, neighbor)]
            if new_distance < d[neighbor]:
                d[neighbor] = new_distance
                parent[neighbor] = source
                heappush(priority_queue, (new_distance, neighbor))
    return d


n, m, s, t = [int(elem) for elem in fin().split(' ')]
adj = defaultdict(list)
yen_weights = dict()
snuuk_weights = dict()
for _ in range(m):
    u, v, a, b = [int(elem) for elem in fin().split(' ')]
    for v1, v2 in [(u, v), (v, u)]:
        adj[v1].append(v2)
        yen_weights[(v1, v2)] = a
        snuuk_weights[(v1, v2)] = b

d_yen = dijkstra_simple(adj, yen_weights, s)
d_snuuk = dijkstra_simple(adj, snuuk_weights, t)
min_cost = float('inf')
min_cost_after_each_year_reversed = []
for vertex in range(n, 0, -1):
    min_cost = min(min_cost, d_yen[vertex] + d_snuuk[vertex])
    min_cost_after_each_year_reversed.append(min_cost)

initial_money = 10**15
print(*(initial_money - cost for cost in reversed(min_cost_after_each_year_reversed)), sep='\n')