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.util.*;
import java.io.*;
public class Main
{
    public static void main(String []args) throws IOException {
    	FastScanner in  = new FastScanner(System.in);
    	PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
    	solve(in, out);
    	in.close();
    	out.close();
    }
    static long gcd(long a, long b){ return (b==0) ? a : gcd(b, a%b); }
    static int gcd(int a, int b){ return (b==0) ? a : gcd(b, a%b); }
    private static int fact(int n) { int ans=1; for(int i=2;i<=n;i++) ans*=i; return ans; }
    static class FastScanner{
	BufferedReader reader;
	StringTokenizer st;
	FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}
	String next(){while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}
	     st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
	String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}
	int    nextInt()   {return Integer.parseInt(next());}
	long   nextLong()  {return Long.parseLong(next());}
	double nextDouble(){return Double.parseDouble(next());}
	char   nextChar()  {return next().charAt(0);}
	int[]  nextIntArray(int n)         {int[] arr= new int[n];   int i=0;while(i<n){arr[i++]=nextInt();}  return arr;}
	long[] nextLongArray(int n)        {long[]arr= new long[n];  int i=0;while(i<n){arr[i++]=nextLong();} return arr;}
	int[]  nextIntArrayOneBased(int n) {int[] arr= new int[n+1]; int i=1;while(i<=n){arr[i++]=nextInt();} return arr;}
	long[] nextLongArrayOneBased(int n){long[]arr= new long[n+1];int i=1;while(i<=n){arr[i++]=nextLong();}return arr;}
	void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}
    }
    
    /********* SOLUTION STARTS HERE ************/
    
    private static void solve(FastScanner in, PrintWriter out){
        int n = in.nextInt();
        int m = in.nextInt();
        int ans = ((n*(n-1))/2)+((m*(m-1))/2);
        out.println(ans);
    }
    
    /************* SOLUTION ENDS HERE **********/
}