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)

# Priority queue with array based heap.
#
# A priority queue is like a standard queue, except that each inserted
# elements is given a certain priority, based on the result of the
# comparison block given at instantiation time. Also, retrieving an element
# from the queue will always return the one with the highest priority
# (see #pop and #top).
#
# The default is to compare the elements in repect to their #<=> method.
# For example, Numeric elements with higher values will have higher
# priorities.
#
# Note that as of version 2.0 the internal queue is kept in the reverse order
# from how it was kept in previous version. If you had used #to_a in the
# past then be sure to adjust for the priorities to be ordered back-to-front
# instead of the oterh way around.
#
class PQueue

  #
  VERSION = "2.1.0"  #:erb: VERSION = "<%= version %>"

  #
  # Returns a new priority queue.
  #
  # If elements are given, build the priority queue with these initial
  # values. The elements object must respond to #to_a.
  #
  # If a block is given, it will be used to determine the priority between
  # the elements. The block must must take two arguments and return `1`, `0`,
  # or `-1` or `true`, `nil` or `false. It should return `0` or `nil` if the
  # two arguments are considered equal, return `1` or `true` if the first
  # argument is considered greater than the later, and `-1` or `false` if
  # the later is considred to be greater than the first.
  #
  # By default, the priority queue retrieves maximum elements first
  # using the #<=> method.
  #
  def initialize(elements=nil, &block) # :yields: a, b
    @que = []
    @cmp = block || lambda{ |a,b| a <=> b }
    replace(elements) if elements
  end

 protected

  #
  # The underlying heap.
  #
  attr_reader :que #:nodoc:

 public

  #
  # Priority comparison procedure.
  #
  attr_reader :cmp

  #
  # Returns the size of the queue.
  #
  def size
    @que.size
  end

  #
  # Alias of size.
  #
  alias length size

  #
  # Add an element in the priority queue.
  #
  def push(v)
    @que << v
    reheap(@que.size-1)
    self
  end

  #
  # Traditional alias for #push.
  #
  alias enq push

  #
  # Alias of #push.
  #
  alias :<< :push

  #
  # Get the element with the highest priority and remove it from
  # the queue.
  #
  # The highest priority is determined by the block given at instantiation
  # time.
  #
  # The deletion time is O(log n), with n is the size of the queue.
  #
  # Return nil if the queue is empty.
  #
  def pop
    return nil if empty?
    @que.pop
  end

  #
  # Traditional alias for #pop.
  #
  alias deq pop

  # Get the element with the lowest priority and remove it from
  # the queue.
  #
  # The lowest priority is determined by the block given at instantiation
  # time.
  #
  # The deletion time is O(log n), with n is the size of the queue.
  #
  # Return nil if the queue is empty.
  #
  def shift
    return nil if empty?
    @que.shift
  end

  #
  # Returns the element with the highest priority, but
  # does not remove it from the queue.
  #
  def top
    return nil if empty?
    return @que.last
  end

  #
  # Traditional alias for #top.
  #
  alias peek top

  #
  # Returns the element with the lowest priority, but
  # does not remove it from the queue.
  #
  def bottom
    return nil if empty?
    return @que.first
  end

  #
  # Add more than one element at the same time. See #push.
  #
  # The elements object must respond to #to_a, or be a PQueue itself.
  #
  def concat(elements)
    if empty?
      if elements.kind_of?(PQueue)
        initialize_copy(elements)
      else
        replace(elements)
      end
    else
      if elements.kind_of?(PQueue)
        @que.concat(elements.que)
        sort!
      else
        @que.concat(elements.to_a)
        sort!
      end
    end
    return self
  end

  #
  # Alias for #concat.
  #
  alias :merge! :concat

  #
  # Return top n-element as a sorted array.
  #
  def take(n=@size)
    a = []
    n.times{a.push(pop)}
    a
  end

  #
  # Returns true if there is no more elements left in the queue.
  #
  def empty?
    @que.empty?
  end

  #
  # Remove all elements from the priority queue.
  #
  def clear
    @que.clear
    self
  end

  #
  # Replace the content of the heap by the new elements.
  #
  # The elements object must respond to #to_a, or to be
  # a PQueue itself.
  #
  def replace(elements)
    if elements.kind_of?(PQueue)
      initialize_copy(elements)
    else
      @que.replace(elements.to_a)
      sort!
    end
    self
  end

  #
  # Return a sorted array, with highest priority first.
  #
  def to_a
    @que.dup
  end

  #
  # Return true if the given object is present in the queue.
  #
  def include?(element)
    @que.include?(element)
  end

  #
  # Push element onto queue while popping off and returning the next element.
  # This is qquivalent to successively calling #pop and #push(v).
  #
  def swap(v)
    r = pop
    push(v)
    r
  end

  #
  # Iterate over the ordered elements, destructively.
  #
  def each_pop #:yields: popped
    until empty?
      yield pop
    end
    nil
  end

  #
  # Pretty inspection string.
  #
  def inspect
    "<#{self.class}: size=#{size}, top=#{top || "nil"}>"
  end

  #
  # Return true if the queues contain equal elements.
  #
  def ==(other)
    size == other.size && to_a == other.to_a
  end

 private

  #
  #
  #
  def initialize_copy(other)
    @cmp  = other.cmp
    @que  = other.que.dup
    sort!
  end

  #
  # The element at index k will be repositioned to its proper place.
  #
  # This, of course, assumes the queue is already sorted.
  #
  def reheap(k)
    return self if size <= 1

    que = @que

    v = que.delete_at(k)
    i = binary_index(que, v)

    que.insert(i, v)

    @que = que

    return self
  end

  #
  # Sort the queue in accorance to the given comparison procedure.
  #
  def sort!
    @que.sort! do |a,b|
      case @cmp.call(a,b)
      when  0, nil   then  0
      when  1, true  then  1
      when -1, false then -1
      else
        warn "bad comparison procedure in #{self.inspect}"
        0
      end
    end
    self
  end

  #
  # Alias of #sort!
  #
  alias heapify sort!

  #
  def binary_index(que, target)
    upper = que.size - 1
    lower = 0

    while(upper >= lower) do
      idx  = lower + (upper - lower) / 2
      comp = @cmp.call(target, que[idx])

      case comp
      when 0, nil
        return idx
      when 1, true
        lower = idx + 1
      when -1, false
        upper = idx - 1
      else
      end
    end
    lower
  end

end # class PQueue

require 'pp'
# IS_DEBUG = true
IS_DEBUG = false
def dputs str
	if(IS_DEBUG)
		puts str
	end
end
def dpp str
	if(IS_DEBUG)
		pp str
	end
end

require 'pp'

INT_MAX = 999999999999999999999999

n,m,s,t=gets.chomp.split(' ').map{|n| n.to_i}
s -= 1
t -= 1
ey = Array.new(n){[]}
es = Array.new(n){[]}
1.upto(m){
	p=gets.chomp.split(' ').map{|n| n.to_i}
	p[0] -= 1
	p[1] -= 1
	ey[p[0]].push([p[1],p[2]])
	ey[p[1]].push([p[0],p[2]])
	es[p[0]].push([p[1],p[3]])
	es[p[1]].push([p[0],p[3]])
}
dpp ey
dpp es

# note : length of from point is set __1__
def dijkstra(e,s,t,darr)
	darr[s] = 0
	pq = PQueue.new([[s,0]]){|a,b| b[1] <=> a[1]}
	while(! pq.empty?)
		u = pq.pop
		# dpp u
		# dpp "poped u #{u} #{u[0][0]}"
		eu = e[u[0]]
		eu.each{|e_tgt|
			d_tgt = darr[e_tgt[0]]
			# dputs "u,e_tgt1 : #{u}, #{e_tgt[1]}"
			# dpp darr
			d_cand = darr[u[0]]+e_tgt[1]
			if(d_tgt.nil? || d_tgt > d_cand)
				d_tgt = darr[e_tgt[0]] = d_cand
				pq.push([e_tgt[0], d_tgt])
			end
		}
	end
end

dysarr = Array.new(n) { nil }
dstarr = Array.new(n) { nil }
dijkstra(ey,s,t,dysarr)
dijkstra(es,t,s,dstarr)
dpp dysarr
dpp "dstarr"
dpp dstarr
minarr = Array.new(n) { |i|  }
0.upto(n-1){|i|
	minarr[i] = dysarr[i]+dstarr[i]
}
minarrsort = minarr.each_with_index.sort{|(d,i),(d2,i2)| d <=> d2}

dpp minarrsort
0.upto(n-1){|yeari|
	minarrsort.each{|d,minarri|
		if(yeari<minarri+1)
			puts 10**15-d
			break
		end
	}
}