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++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <fstream>
#include <string.h>
#pragma GCC optimize("Ofast") //高速化
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//小数点計算
using namespace std;
//strtok(str, ",");  //","で分解
#define prf(f) printf("%f\n", f) //double
#define prc(c) printf("%c\n", c) //char
#define prs(s) printf("%s\n", n) //string
#define prd(d) printf("%d\n", d) //int
#define gno cin.ignore() //一文字無視
#define rep(i, n) for(int i = 0; i < n; i++)
#define reps(i, f, n) for(int i = f; i < n; i++)
#define ch(s1, N) (s1[0]+N) //文字列入れ替え
#define gl(s1) getline(cin,s1) //一列読み込み（一文字無視を使用する）
int main(){
	//cinの高速化
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
		int a , b, c;
		cin >> a >> b;
		c = a * (a-1) /2;
		c = c + b * (b-1) /2;
		cout << c;
		return 0;
}