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 Julia (0.5.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function hpush(h::Array{Tuple{Int,Int},1},x::Tuple{Int,Int})
	push!(h,x)
	if length(h) > 1
		l = length(h)
		k = div(l,2)
		while h[k][2]>x[2]&&k>=1
			tmp = h[k]
			h[k] = x
			h[l] = tmp
			k = max(div(k,2),1)
			l = div(l,2)
		end
	end
end

function hpop(h::Array{Tuple{Int,Int},1})
	x = h[1]
	z = pop!(h)
	if !isempty(h)
		h[1] = z
		k = 1
		l = length(h)
		while k <= l
			if 2*k>=l
				break
			elseif 2*k+1>=l
				if h[2*k][2]>=z[2]
					break
				else
					tmp = h[2*k]
					h[2*k] = z
					h[k] = tmp
					k = 2*k
				end
			else
				if h[2*k][2]>=z[2]&&h[2*k+1][2]>=z[2]
					break
				else
					t = h[2*k][2]<h[2*k+1][2]?2*k:2*k+1
					tmp = h[t]
					h[t] = z
					h[k] = tmp
					k = t
				end
			end
		end
	end
	x
end

function dijkstra(e::Array{Array{Tuple{Int,Int},1},1},k)
	dis = -ones(Int, length(e))
	h = Array{Tuple{Int,Int},1}()
	push!(h,(k,0))
	while !isempty(h)
		v,w = hpop(h)
		dis[v] = dis[v]<0?w:min(dis[v],w)
		for i in 1:length(e[v])
			if dis[e[v][i][1]]<0　|| dis[e[v][i][1]]>w+1
				dis[e[v][i][1]] = w+e[v][i][2]
				hpush(h,(e[v][i][1],w+e[v][i][2]))
			end
		end
	end
	dis
end

function main()
	n,m,s,t = readline() |> split |> parseMap
	e = [Tuple{Int,Int}[] for i in 1:n]
	f = [Tuple{Int,Int}[] for i in 1:n]
	for i in 1:m
		u,v,a,b = readline() |> split |> parseMap
		push!(e[u],(v,a))
		push!(e[v],(u,a))
		push!(f[u],(v,b))
		push!(f[v],(u,b))
	end
	dy = dijkstra(e,s)
	ds = dijkstra(f,t)
	x = Array{Tuple{Int,Int},1}(n)
	for i in 1:n
		x[i] = (i,dy[i]+ds[i])
	end
	x = sort(x,by=x->x[2])
	println("hoge")
end

main()