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.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;

public class Main {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int N = sc.nextInt();
		int M = sc.nextInt();
		int ans = 0;
		if (N >= 2)
			ans += N * (N - 1) / 2;
		if (M >= 2)
			ans += M * (M - 1) / 2;

		System.out.println(ans);
	}

	//以下、自作ライブラリ

	/*
		int N = sc.nextInt();

		int[] a = new int[N];
		for (int i = 0; i < N; i++)
			a[i] = sc.nextInt();

		String s = sc.next();

		String[] a = new String[N];
		for (int i = 0; i < N; i++)
			a[i] = sc.next();

		int[][] a = new int[N][2];
		for (int i = 0; i < N; i++) {
			a[i][0] = sc.nextInt();
			a[i][1] = sc.nextInt();
		}
		Arrays.sort(a, (x, y) -> Integer.compare(x[0], y[0]));

		List<Integer> list = new ArrayList<>();

		Set<Integer> set = new HashSet<>();

		Map<Integer, Integer> map = new HashMap<>();

		nextLineのときは
		Scanner sc = new Scanner(System.in);
	 */

	//繰り返し二乗法による,べき乗計算で剰余を返す
	public static long modPow(long x, long n, final long mod) {
		long sum = 1;
		while (n > 0) {
			if ((n & 1) == 1) { //ビットが立っていたら
				sum = (sum * x) % mod; //x^2^k1 * x^2^k2 * x^2^k3 * ・・・
			}
			x = (x * x) % mod; //((x^2)^2)^2 ・・・
			n >>= 1;
		}
		return sum;
	}

	//組み合わせの数nCrを計算し剰余を返す
	public static long calcNumOfCombinationMod(long n, int r, final long mod) {
		long num = 1;
		for (int i = 0; i < r; i++) {
			num = (num * (n - i)) % mod;
			num = (num * modPow(i + 1, mod - 2, mod)) % mod;
		}
		return num;

	}

	//組み合わせの数nCrを計算
	public static int calcNumOfCombination(int n, int r) {
		int num = 1;
		for (int i = 1; i <= r; i++) {
			num = num * (n - i + 1) / i;
		}
		return num;
	}

	//階乗
	public static int fact(int n) {
		if (n == 1 || n == 0)
			return 1;
		else
			return n * fact(n - 1);
	}

	//指定した値以上の要素が最初に現れる位置
	//要素の数は+1（配列のズレ)
	public static long lowerBound(int n, long[] x, long y) {
		int left = 0;
		int right = n;
		while (left < right) {
			int mid = (left + right) / 2;
			if (x[mid] < y) {
				left = mid + 1;
			} else {
				right = mid;
			}
		}
		return left;
	}

	//指定した値より大きい要素が最初に現れる位置
	//要素の数は+1（配列のズレ)
	public static long upperBound(int n, long[] x, long y) {
		int left = 0;
		int right = n;
		while (left < right) {
			int mid = (left + right) / 2;
			if (x[mid] <= y) {
				left = mid + 1;
			} else {
				right = mid;
			}
		}
		return left;
	}

	//最大公約数
	public static long gcd(long a, long b) {
		if (a < b)
			return gcd(b, a);
		long c = 0;
		c = a % b;
		if (c == 0)
			return b;
		return gcd(b, c);
	}

	//最小公倍数
	public static long lcm(long m, long n) {
		return m * n / gcd(m, n);
	}

	//素数判定
	public static boolean isPrime(int n) {
		if (n < 2)
			return false;
		if (n == 2)
			return true;
		if (n % 2 == 0)
			return false;

		for (int i = 3; i * i <= n; i += 2) {
			if (n % i == 0)
				return false;
		}
		return true;
	}
}

//以下、高速スキャナークラス
class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
		return hasNextByte();
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
			throw new NumberFormatException();
		return (int) nl;
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}
}