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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;
public class Main {
	static Scanner sc;
	static PrintWriter out;
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
		sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int n,m;
		n = sc.nextInt();m = sc.nextInt();
		long ans = n*(n-1)/2 + m*(m-1)/2;
		out.println(ans);
		out.flush();
	} 
	static class Scanner{
		private BufferedReader br;
		private StringTokenizer st;
		public Scanner(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}
		public String next() throws IOException{
			while(st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
		public int nextInt() throws IOException{
			return Integer.parseInt(next());
		}
		public long nextLong() throws IOException{
			return Long.parseLong(next());
		}
		public double nextDouble() throws IOException{
			return Double.parseDouble(next());
		}
		public int[] inArr(int n) throws Exception{
			int a [] = new int[n];
			st = new StringTokenizer(br.readLine());
			for(int i=0;i<n;i++) {
				a[i] = Integer.parseInt(st.nextToken());
			}
			return a;
		}
	}
}