#会津大学オンラインジャッジ問1196 Bridge Removalに合格したコード一番速度の遅い部類
#堀江伸一
def commitCost(old1,now1)
	d=$cons[old1][now1]
	if 1<$cons[old1].size && 1<$cons[now1].size then
		d*=3
	end
	$cost[old1][now1]=d
	$cost[now1][old1]=d
	$cons[now1].each{|next1,d|
		next if next1==old1
		commitCost(now1,next1)
	}
end
 
def calcAns(old1,now1,cost1)
	res=cost1
	$cost[now1].each{|next1,cost2|
		next if old1==next1
		if now1==next1 then
			res=[res,cost1].min
		else
			cost3=$cons[now1][next1]*2
			cost4=cost1-cost2+cost3
			res=[res,cost4].min
			t=calcAns(now1,next1,cost4)
			res=[res,t].min
		end
	}
	return res
end
 
def f(n)
	$cons={}
	$root={}
	$cost={}
	xs=gets.split(" ").map{|e| e.to_i}
	ys=gets.split(" ").map{|e| e.to_i}
	1.upto(n){|i|
		$cons[i]={}
		$cost[i]={}
	}
	from=2
	xs.zip(ys).each{|to,d|
		$cons[from][to]=d
		$cons[to][from]=d
		$cost[from][to]=0
		$cost[to][from]=0
		from+=1
	}
	$cons[1].each{|k,v|
		commitCost(1,k)
	}
 
 
	$allCost=0
	$cost.each{|k1,ks|
		$allCost+=ks.values.sum
	}
	1.upto(n){|i|
		$cons[i][i]=0
		$cost[i][i]=0
	}
 
	$allCost/=2
	ans=10**24
	all=[]
	1.upto(n){|i|
		cost1=$allCost
		t=calcAns(i,i,cost1)
		ans=[t,ans].min
	}
	puts ans
end
 
$cons={}
$root={}
$cost={}
$allCost=0
while true
	n=gets.to_i
	break if n==0
	f(n)
end

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NAoxIDIgMwoxMCAyMCAzMAoxMAoxIDIgMiAxIDUgNSAxIDggOAoxMCAxIDEgMjAgMSAxIDMwIDEgMQozCjEgMQoxIDEKMA==

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