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>
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rrep(i, n) for(int i = ((int)(n)-1); i >= 0; i--)
#define all(x) (x).begin(),(x).end()
 
using namespace std;
using ll = long long;
 
struct edge { int to, cost; };
 
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> P;
 
const int INF = 1e9;
const int MOD = 1e9+7;
const double EPS = 1.0e-10;
const double PI = acos(-1.0);
 
const int mx4[] = {0,1,0,-1};
const int my4[] = {1,0,-1,0};
const int mx8[] = {0,1,1,1,0,-1,-1,-1};
const int my8[] = {1,1,0,-1,-1,-1,0,1};

int main() {
	int n,m;
	cin >> n >> m;
	cout << n*(n-1)/2+m*(m-1)/2 << endl;
	return 0;
}