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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>//入出力
#include<vector>//可変長配列
#include<algorithm>//sort,二分探索,など
#include<bitset>//固定長bit集合
#include<cmath>//pow,logなど
#include<deque>//両端アクセスのキュー
#include<map>//map(辞書)
#include<numeric>//iota(整数列の生成),gcdとlcm(c++17)
#include<queue>//キュー
#include<set>//集合
#include<stack>//スタック
#include<string>//文字列

using namespace std;
typedef long long ll;

#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define RREP(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define REPS(i,a,b) for(ll i=(a);i<(b);i++)
#define RREPS(i,a,b) for(ll i=(a);i>=(b);i--)
#define ALL(x) (x).begin(),(x).end() //sortなどの引数を省略したい
#define SIZE(x) ((ll)(x).size()) //sizeをsize_tからllに直しておく
#define MAX(x) *max_element(ALL(x))
#define INF 1000000000000

int main(){
    int n, m;
    cin >> n >> m;
    cout << (n+m)*(n+m-1)/2 -  n*m << endl;

    return 0;
}