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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.IO;
using System.Diagnostics;

class Myon
{
    public Myon() { }
    public static int Main()
    {
        new Myon().calc();
        return 0;
    }
    Scanner cin;
    Func cin2;
    public static long mod = 1000000007;
    void calc()
    {
        cin = new Scanner();
        cin2 = new Func();
        int N = cin.nextInt();
        int M= cin.nextInt();
        Console.WriteLine((N+M)*(N+M-1)/2-M*N);
    }
}



class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}

class Answer
{
    public void Int2DArrayAns(int[,]x,char c)
    {
        for(int i=0;i<x.GetLength(0);i++)
        {
            for(int j=0;j<x.GetLength(1)-1;j++)
            {
                Console.Write(x[i,j]);
                Console.Write(c);
            }
            Console.WriteLine(x[i,x.GetLength(1)-1]);
        }
    }
    public void Char2DArrayAns(char[,]x,char c)
    {
        for(int i=0;i<x.GetLength(0);i++)
        {
            for(int j=0;j<x.GetLength(1)-1;j++)
            {
                Console.Write(x[i,j]);
                Console.Write(c);
            }
            Console.WriteLine(x[i,x.GetLength(1)-1]);
        }
    }
}

class Func
{
    public static long mod = 1000000007;
    public long GCDLong(long a, long b)
    {
        while(a!=0)
        {
          long t=a;
          a=b%a;
          b= t;
        }
      
        return b;        
    }
    public long mod_pow(long x, long n)
    { 
        long ans = 1;
        while(n > 0)
        {
            if ((n & 1) == 1)
            {
                ans = ans * x;
                ans%=mod;
            }
            x = x*x%mod;
            n >>= 1;
        }
        return ans;
    }
    public long factorial(long x)
    {
        long n=1;
        if(x<=1) return x;
        else
        {
            n *= (x*factorial(x-1));
            n%=mod;
            return n;
        }
    }
    public long combination(long n, long r)
    {
        if ( r * 2 > n ) r = n - r;
		long dividend = 1;
		long divisor = 1;
		for ( long i = 1; i <= r; ++i ) {
			dividend *= ( n - i + 1 );
			dividend %= mod;
 
			divisor *= i;
			divisor %= mod;
		}
 
		return (dividend * mod_pow(divisor,mod - 2)) % mod;   
    }
}