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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

long double factorial(int x){
  long double r;

  r = 1;
  while (0 < x) {
    r = r * x;
    x = x - 1;
  }
  return (r);
}

int main(void){
  int n;
  int m;
  int result;
  long double answer;
  int out;

  result = scanf("%d", &n);
  result = scanf("%d", &m);
  answer = 0;
  if (n < 2 && m < 2)
    answer = 0;
  
  else if ((n < 2 && m == 2) || (n == 2 && m < 2))
    answer = 1;
  else if (n < 2 && m > 2)
    answer = factorial(m) / (factorial(m - 2) * factorial(2));
  else if (n > 2 && m < 2)
    answer = factorial(n) / (factorial(n - 2) * factorial(2));
  else
    answer = factorial(n) / (factorial(n - 2) * factorial(2)) + factorial(m) / (factorial(m - 2) * factorial(2));

  out = (int)round(answer);
  printf("%d", out);
  return (0);
}