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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class Heap
    attr_reader :size
    def initialize
        @heap = []
        @size = 0
    end

    def length
        @size
    end

    def push(x)
        i = @size
        @size += 1
        while i > 0 do
            p = (i - 1) / 2
            break if @heap[p] <= x
            @heap[i] = @heap[p]
            i = p
        end
        @heap[i] = x
        self
    end

    def pop
        return nil if @size == 0
        ret = @heap[0]
        @size -= 1
        x = @heap[@size]
        i = 0
        while i * 2 + 1 < @size do
            a = i * 2 + 1
            b = a + 1
            if b < @size && @heap[b] < @heap[a]
                a = b
            end
            break if @heap[a] >= x
            @heap[i] = @heap[a]
            i = a
        end
        @heap[i] = x
        @heap.pop
        return ret
    end
end

class Dpair
    attr_reader :dist, :vertex
    include Comparable
    def initialize(dist, vertex)
        @dist = dist
        @vertex = vertex
    end
    def <=>(other)
        @dist <=> other.dist
    end
end

# edges[src][dst] = cost
def dijkstra(start, n, edges)
    queue = Heap.new
    dist = Array.new(n+1, Float::INFINITY)
    dist[start] = 0
    queue.push(Dpair.new(0, start))
    while queue.size > 0 do
        dpair = queue.pop
        v = dpair.vertex
        next if dist[v] < dpair.dist
        edges[v].each do |dst, cost|
            if dist[dst] > dist[v] + cost
                dist[dst] = dist[v] + cost
                queue.push(Dpair.new(dist[dst], dst))
            end
        end
    end
    return dist
end

n, m, s, t = gets.split.map(&:to_i)
us, vs, as, bs = [], [], [], []
edge_yen = Array.new(n+1){{}}
edge_snu = Array.new(n+1){{}}
m.times do |i|
    us[i], vs[i], as[i], bs[i] = gets.split.map(&:to_i)
    edge_yen[us[i]][vs[i]] = as[i]
    edge_yen[vs[i]][us[i]] = as[i]
    edge_snu[us[i]][vs[i]] = bs[i]
    edge_snu[vs[i]][us[i]] = bs[i]
end

dist_yen = dijkstra(s, n, edge_yen)
dist_snu = dijkstra(t, n, edge_snu)

ans = Array.new(n+1)
max = 0
FIRST = 10**15
n.downto(1) do |i|
    val = FIRST - (dist_yen[i] + dist_snu[i])
    max = val if max < val
    ans[i] = max
end

(1..n).each do |i|
    puts ans[i]
end