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 (2.7.1) (judge verdict: Accepted)

require 'set'

class Hash
  def push(key, value)
    self[key] = [] if self[key] == nil
    self[key].push(value)
  end
end

class Array
  def lower_bound(value)
    left = -1;
    right = self.length;
    while left + 1 < right
      mid = left + (right - left) / 2;
      if self[mid] >= value
        right = mid
      else
        left = mid
      end
    end
    right
  end

  def unique
    res = [self.first]
    each_cons(2) do |a, b|
      if a != b
        res.push(b)
      end
    end
    res
  end
end

def get_ints
  gets.split.map(&:to_i)
end

def get_ints_minus_one
  get_ints.map { |x| x - 1 }
end

def get_int
  gets.chomp.to_i
end

N, M = get_ints

nx = (N * (N-1)) / 2
mx = (M * (M-1)) / 2

puts nx + mx