Problem:
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Problem F:
There is No Alternative

ICPC (Isles of Coral Park City) consist of several beautiful islands.

The citizens requested construction of bridges between islands to resolve inconveniences of using boats between islands, and they demand that all the islands should be reachable from any other islands via one or more bridges.

The city mayor selected a number of pairs of islands, and ordered a building company to estimate the costs to build bridges between the pairs. With this estimate, the mayor has to decide the set of bridges to build, minimizing the total construction cost.

However, it is difficult for him to select the most cost-efficient set of bridges among those connecting all the islands. For example, three sets of bridges connect all the islands for the Sample Input 1. The bridges in each set are expressed by bold edges in Figure F.1.

Figure F.1. Three sets of bridges connecting all the islands for Sample Input 1

As the first step, he decided to build only those bridges which are contained in all the sets of bridges to connect all the islands and minimize the cost. We refer to such bridges as no alternative bridges. In Figure F.2, no alternative bridges are drawn as thick edges for the Sample Input 1, 2 and 3.

Figure F.2. No alternative bridges for Sample Input 1, 2 and 3

Write a program that advises the mayor which bridges are no alternative bridges for the given input.

Input

The input consists of a single test case.

$N$ $M$

$S_1$ $D_1$ $C_1$

.

.

.

$S_M$ $D_M$ $C_M$

The first line contains two positive integers $N$ and $M$. $N$ represents the number of islands and each island is identified by an integer 1 through $N$. $M$ represents the number of the pairs of islands between which a bridge may be built.

Each line of the next $M$ lines contains three integers $S_i$, $D_i$ and $C_i$ ($1 \leq i \leq M$) which represent that it will cost $C_i$ to build the bridge between islands $S_i$ and $D_i$. You may assume $3 \leq N \leq 500$, $N − 1 \leq M \leq min(50000, N(N − 1)/2)$, $1 \leq S_i < D_i \leq N$, and $1 \leq C_i \leq 10000$. No two bridges connect the same pair of two islands, that is, if $i \ne j$ and $S_i = S_j$, then $D_i \ne D_j$ . If all the candidate bridges are built, all the islands are reachable from any other islands via one or more bridges.

Output

Output two integers, which mean the number of no alternative bridges and the sum of their construction cost, separated by a space.

Sample Input 1

4 4
1 2 3
1 3 3
2 3 3
2 4 3

Sample Output 1

1 3

Sample Input 2

4 4
1 2 3
1 3 5
2 3 3
2 4 3

Sample Output 2

3 9

Sample Input 3

4 4
1 2 3
1 3 1
2 3 3
2 4 3

Sample Output 3

2 4

Sample Input 4

3 3
1 2 1
2 3 1
1 3 1

Sample Output 4

0 0

Solution in Ruby (judge verdict: Time Limit Exceeded)

class UnionFind
	def initialize
		@obj_to_node = {}
	end

	def add(obj)
		@obj_to_node[obj] ||= Node.new(obj)
	end

	def size(x = nil)
		x ? @obj_to_node[x].root.size : @obj_to_node.size
	end

	class Node
		def initialize(obj)
			@obj = obj
			@parent = self
			@size = 1
			@rank = 0
		end

		def root
			return self if @parent == self
			@parent = @parent.root
		end	

		attr_accessor :parent, :size, :rank
	end

	def union(x, y)
		u = (@obj_to_node[x] ||= Node.new(x)).root
		v = (@obj_to_node[y] ||= Node.new(y)).root
		return if u == v
		if u.rank > v.rank
			v.parent = u
			u.size += v.size
		elsif v.rank > u.rank
			u.parent = v
			v.size += u.size
		else
			u.parent = v
			v.rank += 1
			v.size += u.size
		end
	end

	def same?(x, y)
		return false unless u = @obj_to_node[x]
		return false unless v = @obj_to_node[y]
		u.root == v.root
	end
end

require 'set'

class Graph
	def initialize(*options)
		@options = Set[*options]
	end

	def add_node(property = {})
		Node.new(property)
	end

	def add_edge(u, v, property = {})
		e = Edge.new(u, v, property)
		u.add_edge(e)
		v.add_edge(e)
		e
	end
	
	def del_edge(e)
		u, v = e.opposite.keys
		u.del_edge(e)
		v.del_edge(e)
		e
	end

	class Node
		def initialize(property)
			@property = property
			@edges = Set.new
		end

		def to_s
			@property[:index]
		end

		def add_edge(e)
			@edges << e
		end

		def del_edge(e)
			@edges.delete(e)
		end

		def to_s
			@property[:name] || super
		end

		def each_edge
			@edges.each{|e| 
				yield e
			}
		end

		attr_reader :property, :edges
	end

	class Edge
		def initialize(u, v, property)
			@opposite = {u => v, v => u}
			@property = property
		end
		attr_reader :opposite, :property

		def length
			@property[:length] || 1
		end
	end

	def each_out_connection_of(u)
		u.each_edge{|e|
			yield(e.opposite[u], e.length) }
	end

	alias :each_in_connection_of :each_out_connection_of
end

n, m = gets.split.map &:to_i

g = Graph.new
nodes = (0..n).map {|i| g.add_node({:index => i})}
edges = []

(1..m).each do
	i, j, c = gets.split.map &:to_i
	edges << g.add_edge(nodes[i], nodes[j], {:length => c})
end
edges.sort_by!(&:length)

mst = []
uf = UnionFind.new

history = {}

edges.each do |e|
	u, v = e.opposite.keys	
	i, j = u.property[:index], v.property[:index]
	#p [u.property[:index], v.property[:index]]
	next if uf.same?(i, j)
	history[e] = [Marshal.load(Marshal.dump(uf)), mst.map(&:length).inject(0, :+)]
	mst << e
	uf.union(i, j)
end

C = mst.map(&:length).inject(:+)

k = 0
s = 0
mst.each.with_index do |deleted, t0|
	e = nil
	e = edges.shift until e == deleted
	uf, c = history[e]
	t = t0

	edges.each do |e|
		next if e == deleted
		u, v = e.opposite.keys	
		i, j = u.property[:index], v.property[:index]
		next if uf.same?(i, j)
		uf.union(i, j)
		c += e.length
		t += 1
	end

	if c > C || t < n - 1
		k += 1
		s += deleted.length
	end
end

puts [k, s] * ' '