Problem:
Score : 100 points

Problem Statement

We have N+M balls, each of which has an integer written on it.

It is known that:

The numbers written on N of the balls are even.

The numbers written on M of the balls are odd.

Find the number of ways to choose two of the N+M balls (disregarding order) so that the sum of the numbers written on them is even.

It can be shown that this count does not depend on the actual values written on the balls.

Constraints

0 \leq N,M \leq 100

2 \leq N+M

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the answer.

Sample Input 1

2 1

Sample Output 1

1

For example, let us assume that the numbers written on the three balls are 1,2,4.

If we choose the two balls with 1 and 2, the sum is odd;

If we choose the two balls with 1 and 4, the sum is odd;

If we choose the two balls with 2 and 4, the sum is even.

Thus, the answer is 1.

Sample Input 2

4 3

Sample Output 2

9

Sample Input 3

1 1

Sample Output 3

0

Sample Input 4

13 3

Sample Output 4

81

Sample Input 5

0 3

Sample Output 5

3

Solution in Ruby (2.3.3) (judge verdict: Accepted)

class InputLine
  def self.to_s
    gets.chomp
  end

  def self.to_i
    InputLine.to_s.to_i
  end
  
  def self.to_f
    InputLine.to_s.to_f
  end

  def self.to_sa
    InputLine.to_s.split(' ')
  end

  def self.to_ia
    InputLine.to_sa.map(&:to_i)
  end
  
  def self.to_fa
    InputLine.to_sa.map(&:to_f)
  end
end

def calc(x)
  (x * (x - 1)) / 2
end

def main(argv)
  (n, m) = InputLine.to_ia

  puts (calc(n) + calc(m)).to_s
end

if self.to_s == 'main' then
  main(ARGV)
end