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.

Input is given from Standard Input in the following format:

```
$N$ $M$
```

Print the answer.

-   $0 \leq N,M \leq 100$
-   $2 \leq N+M$
-   All values in input are integers.

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 $1$.