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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        A solver = new A();
        solver.solve(1, in, out);
        out.close();
    }

    static class A {
        static final int mod = 1000000007;

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // 入力
            int n = Integer.parseInt(in.next());
            int m = Integer.parseInt(in.next());
            int ans = 0;

            ans = nCk(n, 2, mod) + nCk(m, 2, mod);
            if (n == 0) ans--;
            if (m == 0) ans--;
            if (n == 1) ans--;
            if (m == 1) ans--;

            // 出力
            out.println(ans);

        }

        public static int nCk(int n, int k, int M) {
            long ret = 1;
            int min = Math.min(k, n - k);
            for (int i = 1; i <= min; i++) {
                ret = (ret * pow(i, M - 2, M)) % M;
            }
            for (int i = n - min + 1; i <= n; i++) {
                ret = (ret * i) % M;
            }
            return (int) ret;
        }

        public static long pow(long a, long b, int M) {
            long ret = 1;
            long tmp = a;
            while (b > 0) {
                if ((b & 1) == 1) ret = (ret * tmp) % M;
                tmp = (tmp * tmp) % M;
                b = b >> 1;
            }
            return ret;
        }

    }
}