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 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
#define all(x)(x).begin(), (x).end() // 昇順ソート
#define rall(v)(v).rbegin(), (v).rend() // 降順ソート
#define INF 1 LL << 60
typedef long long int LL;
typedef long long int ll;
#define pll pair < ll, ll >
#define F first
#define S second
const int MOD = 1000000007;
template < class T > bool chmax(T & a,const T & b) { if (a < b) { a = b; return true; } return false; }
template < class T > bool chmin(T & a,const T & b) { if (a > b) { a = b; return true; } return false; }
// 任意のlogは　対数の底の変換を使う    log(N) / log(任意の底)

const int MAX = 100000;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++) {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k) {
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}


int main() {
   int N,M;cin >> N >> M;

    COMinit();
    int a = COM(N,2);
    int b = COM(M,2);

    cout << a+b << endl;
}