Problem:
Problem D: Traveling by Stagecoach

Once upon a time, there was a traveler.

He plans to travel using stagecoaches (horse wagons).
His starting point and destination are fixed,
but he cannot determine his route.
Your job in this problem is to write a program
which determines the route for him.

There are several cities in the country,
and a road network connecting them.
If there is a road between two cities, one can travel by a stagecoach
from one of them to the other.
A coach ticket is needed for a coach ride.
The number of horses is specified in each of the tickets.
Of course, with more horses, the coach runs faster.

At the starting point, the traveler has a number of coach tickets.
By considering these tickets and the information on the road network,
you should find the best possible route that takes him to the destination
in the shortest time.
The usage of coach tickets should be taken into account.

The following conditions are assumed.

A coach ride takes the traveler from one city to another
directly connected by a road.
In other words, on each arrival to a city, he must change the coach.

Only one ticket can be used for a coach ride between two cities
directly connected by a road.

Each ticket can be used only once.

The time needed for a coach ride is the distance between two cities
divided by the number of horses.

The time needed for the coach change should be ignored.

Input

The input consists of multiple datasets, each in the following format.
The last dataset is followed by a line containing five zeros
(separated by a space).

n m p a b

t1 t2 ... tn

x1 y1 z1

x2 y2 z2

...

xp yp zp

Every input item in a dataset is a non-negative integer.
If a line contains two or more input items, they are separated by a space.

n is the number of coach tickets.
You can assume that the number of tickets is between 1 and 8.
m is the number of cities in the network.
You can assume that the number of cities is between 2 and 30.
p is the number of roads between cities, which may be zero.

a is the city index of the starting city.
b is the city index of the destination city.
a is not equal to b.
You can assume that all city indices in a dataset (including the above two)
are between 1 and m.

The second line of a dataset gives the details of coach tickets.
ti is the number of horses specified in the i-th
coach ticket (1<=i<=n).
You can assume that the number of horses is between 1 and 10.

The following p lines give the details of roads between cities.
The i-th road connects two cities with city indices
xi and yi,
and has a distance zi (1<=i<=p).
You can assume that the distance is between 1 and 100.

No two roads connect the same pair of cities.
A road never connects a city with itself.
Each road can be traveled in both directions.

Output

For each dataset in the input, one line should be output as specified below.
An output line should not contain extra characters such as spaces.

If the traveler can reach the destination, the time needed
for the best route (a route with the shortest time) should be printed.
The answer should not have an error greater than 0.001.
You may output any number of digits after the decimal point,
provided that the above accuracy condition is satisfied.

If the traveler cannot reach the destination,
the string "Impossible" should be printed.
One cannot reach the destination either when there are no routes
leading to the destination, or when the number of tickets is not sufficient.
Note that the first letter of "Impossible" is in uppercase,
while the other letters are in lowercase.

Sample Input

3 4 3 1 4
3 1 2
1 2 10
2 3 30
3 4 20
2 4 4 2 1
3 1
2 3 3
1 3 3
4 1 2
4 2 5
2 4 3 4 1
5 5
1 2 10
2 3 10
3 4 10
1 2 0 1 2
1
8 5 10 1 5
2 7 1 8 4 5 6 3
1 2 5
2 3 4
3 4 7
4 5 3
1 3 25
2 4 23
3 5 22
1 4 45
2 5 51
1 5 99
0 0 0 0 0

Output for the Sample Input

30.000
3.667
Impossible
Impossible
2.856

Solution in Ruby (judge verdict: Accepted)

class Heap
	require 'forwardable'
	extend Forwardable

	def initialize
		@array = []
		@index = {}
	end

	def add(obj, key)
		@array << [obj, key]
		@index[obj] = @array.size - 1
		lift_up(@array.size - 1)
	end

	def swap(i, j)
		u = @array[i]
		v = @array[j]
		@index[u[0]], @index[v[0]] = j, i
		@array[i], @array[j] = v, u
	end

	def lift_up(i)
		return if i == 0
		j = (i - 1) / 2
		if @array[i][1] < @array[j][1]
			swap(i, j)
			lift_up(j)
		end
	end

	def del_min
		raise "heap is empty!" if empty?
		min = @array[0]
		if size > 1
			swap(0, @array.size-1)
			@array.pop
			lift_down(0)
		else
			@array.pop
		end
		@index.delete(min[0])
		min
	end

	def inc_key(obj, diff)
		i = @index[obj]
		@array[i][1] += diff
		lift_down(i)
	end

	def dec_key(obj, diff)
		i = @index[obj]
		@array[i][1] -= diff
		lift_up(i)
	end

	def update_key(obj, new_key)
		i = @index[obj]
		return unless i
		return if new_key >= @array[i][1]
		@array[i][1] = new_key
		lift_up(i)
	end

	def min_obj
		@array[0][0]
	end

	def min_key
		@array[0][1]
	end

	def lift_down(i)
		j = 2 * i + 1
		k = j + 1
		if k >= size
			return if j >= size
			if @array[i][1] > @array[j][1]
				swap(i, j)
				lift_down(j)
			end
		else
			l = (@array[j][1] < @array[k][1] ? j : k)
			if @array[i][1] > @array[l][1]
				swap(i, l)
				lift_down(l)
			end
		end
	end

	def_delegators :@array, :empty?, :size
end

module Graphical
	require 'set'

	def reachable_from(s, max_depth = nil)
		reached = Set[s]
		queue = [s]
		depth = 0
		loop {
			break if max_depth && depth >= max_depth
			next_queue = []
			until queue.empty?
				u = queue.shift
				each_to_node_of(u){|v|
					next if reached.include?(v)
					reached << v
					next_queue << v
				}
			end
			break if next_queue.empty?
			queue = next_queue
			depth += 1
		}
		return reached
	end

	def dijkstra(start, setting = {})
		raise "invalid setting" if setting.class != Hash
		raise "invalid setting" if setting.keys - [:goal, :multi_source, :multi_sink, :max_dist] != []


		
		start = Set[start] if !setting[:multi_source]
		goal = setting[:goal]
		goal = Set[goal] if goal && !setting[:multi_sink]
		max_dist = setting[:max_dist]

		h = Heap.new
		reached = Set.new
		start.each{|v|
			h.add(v, 0)
			reached << v
		}

		dist = {}
		until h.empty?
			u, key = h.del_min

			return key if setting[:goal] && goal.include?(u)
			break if max_dist && key > max_dist
			dist[u] = key
			each_to_node_of(u){|v|
				d = dist[u] + edge_length(u, v)
				if reached.include?(v)
					h.update_key(v, d)
				else
					h.add(v, d)
					reached << v
				end
			}
		end
		goal ? nil : dist
	end

	def bidir_dijkstra(s, t)
		f_dist = {}
		b_dist = {}

		f_heap = Heap.new
		b_heap = Heap.new

		f_heap.add(s, 0)
		b_heap.add(t, 0)

		f_reached = Set[s]
		b_reached = Set[t]

		f_current_dist = b_current_dist = 0
		best = Float::INFINITY

		loop {
			if !f_heap.empty?
				u, key = f_heap.del_min
				f_dist[u] = key
				f_current_dist = key
				each_to_node_of(u){|v|
					d = f_dist[u] + edge_length(u, v)
					if b_dist[v]
						path_len = d + b_dist[v]
						best = path_len if path_len < best
					end
					if f_reached.include?(v)
						f_heap.update_key(v, d)
					else
						f_heap.add(v, d)
						f_reached << v
					end
				}
			end

			if !b_heap.empty?
				u, key = b_heap.del_min
				b_dist[u] = key
				b_current_dist = key
				each_from_node_of(u){|v|
					d = b_dist[u] + edge_length(v, u)
					if f_dist[v]
						path_len = d + f_dist[v]
						best = path_len if path_len < best
					end
					if b_reached.include?(v)
						b_heap.update_key(v, d)
					else
						b_heap.add(v, d)
						b_reached << v
					end
				}
			end

			return best if f_current_dist + b_current_dist >= best
			return nil if f_heap.empty? && b_heap.empty?
		}
	end

	def prim(start)
		h = Heap.new
		h.add(start, 0)
		reached = Set[start]
		tree_len = 0

		until h.empty?
			u, key = h.del_min
			tree_len += key

			each_to_node_of(u){|v|
				d = edge_length(u, v)
				if reached.include?(v)
					h.update_key(v, d)
				else
					h.add(v, d)
					reached << v
				end
			}
		end
		tree_len
	end
end

class Graph
	include Graphical
	require 'set'

	def add_node(label = nil)
		Node.new(label)
	end

	def add_edge(u, v, label = nil)
		e = Edge.new(u, v, label)
		u.neighbour[v] = e
		v.neighbour[u] = e
		e
	end
	
	def del_edge(e)
		u, v = e.from, e.to
		u.neighbour.delete(v)
		v.neighbour.delete(u)
		e
	end

	class Node
		def initialize(label)
			@label = label
			@neighbour = {}
		end

		def to_s
			@label
		end
		
		attr_accessor :neighbour
		attr_reader :label
	end

	class Edge
		def initialize(u, v, label = nil)
			@from = u	
			@to = v
			@label = label
		end
		attr_reader :from, :to, :label
	end


	def each_to_node_of(u)
		u.neighbour.each_key{|v|
			yield v
		}
	end
	alias :each_from_node_of :each_to_node_of

	def edge_length(u, v)
		u.neighbour[v] ? 1 : nil
	end

end

class WeightedGraph < Graph
	def add_edge(u, v, length, label = nil)
		e = Edge.new(u, v, length, label)
		u.neighbour[v] = e
		v.neighbour[u] = e
		e
	end

	class Edge
		def initialize(u, v, length, label = nil)
			@from = u	
			@to = v
			@length = length
			@label = label
		end
		attr_reader :from, :to, :length, :label
	end

	def edge_length(u, v)
		u.neighbour[v].length
	end
end

extend Graphical
	def each_to_node_of(x)
		u, rest_tickets = x
		$g.each_to_node_of(u){|v|
			(0..rest_tickets.size-1).each{|i|
				copy = rest_tickets.dup
				copy.delete_at(i)
				yield [v, copy]
			}
		}
	end

	def edge_length(u, v)
		l = u[0].neighbour[v[0]].length
		raise "hoge" if u[1].size <= v[1].size
		(0..v[1].size-1).each{|i|
			return l / u[1][i] if u[1][i] != v[1][i]
		}
		l / u[1][-1]
	end

loop {
	t, n, m, a, b = gets.split.map &:to_i	
	break if t == 0

	$g = WeightedGraph.new
	nodes = (0..n).map{|i| $g.add_node(i) }
	tickets = gets.split.map(&:to_i).sort

	m.times{
		i, j, d = gets.split.map &:to_i
		$g.add_edge(nodes[i], nodes[j], d.to_f)
	}

	goals = (0..tickets.size).map{|k| tickets.combination(k).to_a}.flatten(1).uniq.map{|ts| [nodes[b], ts]}
	d = dijkstra([nodes[a], tickets], {:multi_sink => true, :goal => goals })
	if d
		printf("%.5f\n", d)
	else
		puts "Impossible"
	end
}