Problem:
Score : 100 points

Problem Statement

There is a right triangle ABC with ∠ABC=90°.

Given the lengths of the three sides, |AB|,|BC| and |CA|, find the area of the right triangle ABC.

It is guaranteed that the area of the triangle ABC is an integer.

Constraints

1 \leq |AB|,|BC|,|CA| \leq 100

All values in input are integers.

The area of the triangle ABC is an integer.

Input

Input is given from Standard Input in the following format:

|AB| |BC| |CA|

Output

Print the area of the triangle ABC.

Sample Input 1

3 4 5

Sample Output 1

6

This triangle has an area of 6.

Sample Input 2

5 12 13

Sample Output 2

30

This triangle has an area of 30.

Sample Input 3

45 28 53

Sample Output 3

630

This triangle has an area of 630.

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

begin
input = STDIN.readline.split()
n = input[0].to_i
k = input[1].to_i
td_list = STDIN.readlines
td_list.map!{|line|line.split()}
hash_list = []
for td in td_list do
    t = td[0]
    d = td[1].to_i
    hash_list.push(Hash[:t, t, :d, d])
end
hash_list.sort! do |a, b|
    a[:d] <=> b[:d]
end
hash_list.reverse!
answer = hash_list[Range.new(0, k, true)]
counter = Hash.new(0)
for td_a in answer do
    counter[td_a[:t]] += 1
end
once = answer.select do |once_td|
    counter[once_td[:t]] == 1
end
twice = answer.select do |twice_td|
    counter[twice_td[:t]] > 1
end
twice.reverse!
rest = hash_list[Range.new(k, n, true)].select do |rest_td|
    counter[rest_td[:t]] == 0
end
max = 0
for a in answer do
    max += a[:d]
end
counts = counter.size
max += (counts * counts)
twice.each_index do |i|
    if rest.size > 0 && counter[twice[i][:t]] > 0 then
        counter[twice[i][:t]] -= 1
        p = rest.shift
        counter[p[:t]] += 1
        twice[i][:t] = p[:t]
        twice[i][:d] = p[:d]
        counts += 1
        sum = (once + twice).map{|ot_td|ot_td[:d]}.sum + (counts * counts)
        if max < sum then 
            max = sum
        end
    end
end

puts(max)
rescue
    puts(0)
end