Problem:
Score : 1100 points

Problem Statement

Coloring of the vertices of a tree G is called a good coloring when, for every pair of two vertices u and v painted in the same color, picking u as the root and picking v as the root would result in isomorphic rooted trees.

Also, the colorfulness of G is defined as the minimum possible number of different colors used in a good coloring of G.

You are given a tree with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. We will construct a new tree T by repeating the following operation on this tree some number of times:

Add a new vertex to the tree by connecting it to one of the vertices in the current tree with an edge.

Find the minimum possible colorfulness of T.
Additionally, print the minimum number of leaves (vertices with degree 1) in a tree T that achieves the minimum colorfulness.

Notes

The phrase "picking u as the root and picking v as the root would result in isomorphic rooted trees" for a tree G means that there exists a bijective function f_{uv} from the vertex set of G to itself such that both of the following conditions are met:

f_{uv}(u)=v

For every pair of two vertices (a,b), edge (a,b) exists if and only if edge (f_{uv}(a),f_{uv}(b)) exists.

Constraints

2 \leq N \leq 100

1 \leq a_i,b_i \leq N(1\leq i\leq N-1)

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}

Output

Print two integers with a space in between.
First, print the minimum possible colorfulness of a tree T that can be constructed.
Second, print the minimum number of leaves in a tree that achieves it.

It can be shown that, under the constraints of this problem, the values that should be printed fit into 64-bit signed integers.

Sample Input 1

5
1 2
2 3
3 4
3 5

Sample Output 1

2 4

If we connect a new vertex 6 to vertex 2, painting the vertices (1,4,5,6) red and painting the vertices (2,3) blue is a good coloring.
Since painting all the vertices in a single color is not a good coloring, we can see that the colorfulness of this tree is 2.
This is actually the optimal solution. There are four leaves, so we should print 2 and 4.

Sample Input 2

8
1 2
2 3
4 3
5 4
6 7
6 8
3 6

Sample Output 2

3 4

Sample Input 3

10
1 2
2 3
3 4
4 5
5 6
6 7
3 8
5 9
3 10

Sample Output 3

4 6

Sample Input 4

13
5 6
6 4
2 8
4 7
8 9
3 2
10 4
11 10
2 4
13 10
1 8
12 1

Sample Output 4

4 12

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

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'

# from s(s) to t with limited depth d of graph e
# s is assumed to be an array [s1] or [s1,s2]
def count(g,s,d)
	arrsize = (d+2)/2
	maxarr = Array.new(arrsize){0}
	def makearr(g,s,nott,maxarr)
		stack = [] # stack
		f = Array.new(g.size){false} # is searched flag
		# tr = Array.new(g.size){nil}
		# init
		dmax = 0
		emax = s # e of dmax
		f[s]=true
		enext = E[s].reject{|e| e == nott}
		# evaluation
		if(enext.size > maxarr[0])
			maxarr[0] = enext.size
		end
		# next edge push
		enext.each{|v|
			# tr[v]=s
			stack.push([v,1]) #v,d to v
		} 
		while(!stack.empty?)
			curv,curd = stack.pop
			dputs "(curv,curd) = (#{curv},#{curd})"
			f[curv]=true
			enext = E[curv].reject{|e| f[e]}
			# evaluation
			if(enext.size > maxarr[curd])
				maxarr[curd] = enext.size
			end
			# next edge push
			enext.each{|v|
				# tr[v]=curv
				stack.push([v,curd+1])
			}
		end
	end
	case s.size
	when 1
		makearr(g,s[0],s[0],maxarr)
	when 2
		dputs "first makearr"
		makearr(g,s[0],s[1],maxarr)
		dputs "second makearr"
		makearr(g,s[1],s[0],maxarr)
	end
	maxarr[-1]=1
	dpp maxarr
	return s.size*maxarr.inject(1){|prod,n| prod*n}
end
# returns diameter and central edge(s)
def diameter(g)
	def farthest(g,s)
		stack = [] # stack
		f = Array.new(g.size){false} # is searched flag
		tr = Array.new(g.size){nil}
		# init
		dmax = 0
		emax = s # e of dmax
		tr[s]=s
		f[s]=true
		E[s].each{|v|
			tr[v]=s
			stack.push([v,1]) #v,d to v
		} 
		while(!stack.empty?)
			curv,curd = stack.pop
			f[curv]=true
			if(dmax < curd)
				dmax = curd
				emax = curv
			end
			dputs "(curv,curd) = (#{curv},#{curd})"
			E[curv].each{|v|
				if !f[v]
					tr[v]=curv
					stack.push([v,curd+1])
				end
			}
		end
		return [dmax,emax,tr]
	end
	# diameter by two execute
	d,e,tr = farthest(g,0)
	dputs "d = #{d}"
	dputs "e = #{e}"
	dpp tr
	d,e,tr = farthest(g,e)
	dputs "d = #{d}"
	dputs "e = #{e}"
	dpp tr
	# trace 
	etr = e
	len = 0
	centd = d/2
	offs = d%2
	cente = []
	while(true)
		etr = tr[etr]
		len += 1
		if(len == centd)
			cente << etr
		elsif(len == centd+offs)
			cente << etr
			break
		elsif(len > centd+offs)
			break
		end
	end
	return [d, cente]
end

N,_=gets.chomp.split(' ').map{|n| n.to_i}
if(N==2)
	puts "1 2"
	exit(0)
end
E = Array.new(N){[]}
1.upto(N-1){
	p=gets.chomp.split(' ').map{|n| n.to_i}
	p[0]-=1
	p[1]-=1
	E[p[0]].push(p[1])
	E[p[1]].push(p[0])
}
d,s = diameter(E)
dpp s
mind = count(E,s,d)
if s.size == 1
	# center of the generated tree is the center or the center + neibor
	E[s[0]].each{|sn|
		tmp = count(E,[s[0],sn],d)
		dputs "count #{s[0]},#{sn} is #{tmp}"
		mind = tmp if tmp < mind
	}
end
puts "#{(d+2)/2} #{mind}"