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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*
import java.io.PrintWriter

val pw = PrintWriter(System.out)
val MOD = 1000000007L
val INF = 2147483647
val LINF = 9223372036854775807L

fun main(args: Array<String>){
    solve()
    pw.flush()    
}

fun solve(){
    val (n, m) = nextIntList()
    println(n * (n - 1) / 2 + m * (m - 1) / 2)


}

// Print
fun println(v: String){
    pw.println(v)
}
fun print(v: String){
    pw.print(v)
}

// Read
fun next() = readLine()!!
fun nextInt() = next().toInt()
fun nextLong() = next().toLong()
fun nextDouble() = next().toDouble()
fun nextList() = next().split(" ")
fun nextIntList() = next().split(" ").map{ it.toInt() }
fun nextLongList() = next().split(" ").map{ it.toLong() }
fun nextDoubleList() = next().split(" ").map{ it.toDouble() }
fun nextAryln(n: Int) = Array(n){ next() }
fun nextIntAryln(n: Int) = IntArray(n){ nextInt() }
fun nextLongAryln(n: Int) = LongArray(n){ nextLong() }
fun nextDoubleAryln(n: Int) = DoubleArray(n) { nextDouble() }

// Math
fun abs(n: Int) : Int = Math.abs(n)
fun abs(n: Long) : Long = Math.abs(n)
fun abs(n: Double) : Double = Math.abs(n)
fun max(a: Int = -INF, b: Int = -INF, c: Int = -INF, d: Int = -INF, e: Int = -INF): Int = listOf(a, b, c, d, e).max()!!
fun max(a: Long = -LINF, b: Long = -LINF, c: Long = -LINF, d: Long = -LINF, e: Long = -LINF): Long = listOf(a, b, c, d, e).max()!!.toLong()
fun min(a: Int = INF, b: Int = INF, c: Int = INF, d: Int = INF, e: Int = INF): Int = listOf(a, b, c, d, e).min()!!
fun min(a: Long = LINF, b: Long = LINF, c: Long = LINF, d: Long = LINF, e: Long = LINF): Long = listOf(a, b, c, d, e).min()!!.toLong()
fun prime(from: Long, to: Long = from) : List<Long>{
    return (from..to).filter{ i ->
        val max = Math.sqrt(i.toDouble()).toLong()
        (2..max).all{ j -> i % j != 0L}
    }
}
fun gcd(a: Long, b: Long) : Long = if(a % b == 0L) b else gcd(b, (a % b))
fun lcm(a: Long, b: Long) : Long = a / gcd(a, b) * b
fun modpow(a: Long, n: Long, p:Long = MOD) : Long {
    var res = 1L
    var ar = a
    var nr = n
    while(nr > 0){
        if((nr and 1) == 1L) res = res * ar % p
        ar = ar * ar % p
        nr = nr shr 1
    }
    return res
}
fun modinv(a: Long, p: Long = MOD) : Long = modpow(a, p - 2, p)
fun ncr(n: Long, r: Long) : Long {
    var a = 1L
    var b = 1L
    for (i in 1..r) {
        a = a * (n + 1 - i) % MOD
        b = b * i % MOD
    }
    return modinv(b, MOD) * a % MOD
}