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: Runtime Error)

#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int dx[4] = {-1, 0, 0, 1};
const int dy[4] = {0, -1, 1, 0};

// Self settings
// clang-format off
#define MAX_N 100000
#define MAX 100000
#define INFTY (1<<30)
#define EPS (1e-10)
#define equals(a, b) (fabs((a) - (b)) < EPS)
#define REP(i, N) for (int i = 0; i < (int)(N); ++i)
#define SLN(i,N) (i == N-1 ? "\n" : " ")
ll fact(ll n) { ll res = 1; for(ll i=2;i<=n;++i) res = res * i; return res;}
ll gcd(ll a,ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const ll MOD = 1e9+7;
const ll INF = 1LL << 60;
const int inf = 1000100011;
class Point { public: double x, y; Point(double x = 0, double y = 0) : x(x), y(y) {} Point operator+(Point p) { return Point(x + p.x, y + p.y); } Point operator-(Point p) { return Point(x - p.x, y - p.y); } Point operator*(double a) { return Point(a * x, a * y); } Point operator/(double a) { return Point(x / a, y / a); } double abs() { return sqrt(norm()); } double norm() { return x * x + y * y; } bool operator<(const Point& p) const { return x != p.x ? x < p.x : y < p.y; } bool operator==(const Point& p) const { return fabs(x - p.y) < EPS && fabs(y - p.y) < EPS; } };
typedef Point Vector;
double norm(Vector a) { return a.x * a.x + a.y * a.y; }
double abs(Vector a) { return sqrt(norm(a));}
double dot(Vector a, Vector b) { return a.x * b.x + a.y * b.y; }
double cross(Vector a, Vector b) { return a.x * b.y - a.y * b.x; }
struct Segment { Point p1, p2; };
typedef Segment Line;
// clang-format on

int nCr(int n, int c) {
  ll a = 1, b = 1, d = 1;
  for (int i = 1; i <= n; i++) {
    a *= i;
  }
  for (int i = 1; i <= c; i++) {
    b *= i;
  }
  for (int i = 1; i <= n - c; i++) {
    d *= i;
  }
  return a / (b * d);
}

int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, M;
  cin >> N >> M;

  ll ans = nCr(N, 2) + nCr(M, 2);
  cout << ans << endl;

  return 0;
}