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)

// template version 1.15
using namespace std;
#include <bits/stdc++.h>

// varibable settings
#define int long long
const int INF=1e18;

// define basic macro {{{
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define _rrep(i,n) rrepi(i,0,n)
#define rrepi(i,a,b) for(int i=(int)((b)-1);i>=(int)(a);--i)
#define rrep(...) _overload3(__VA_ARGS__,rrepi,_rrep,)(__VA_ARGS__)
#define each(i,a) for (auto&& i : a)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define mt(a, b, c) make_tuple(a, b, c)
#define ub upper_bound
#define lb lower_bound
#define posl(A, x) (lower_bound(all(A), x)-A.begin())
#define posu(A, x) (upper_bound(all(A),x)-A.begin())
template<class T> inline void chmax(T &a, const T &b) { if((a) < (b)) (a) = (b); }
template<class T> inline void chmin(T &a, const T &b) { if((a) > (b)) (a) = (b); }

#define divceil(a,b) ((a)+(b)-1)/(b)
#define is_in(x, a, b) ((a)<=(x) && (x)<(b))
#define uni(x) sort(all(x));x.erase(unique(all(x)),x.end())
#define slice(l, r) substr(l, r-l)

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long double ld;
typedef pair<int,int> pii;
typedef tuple<int,int,int> iii;

template<typename T> using PQ = priority_queue<T, vector<T>, greater<T>>;
struct Fast { Fast(){ std::cin.tie(0); ios::sync_with_stdio(false); } } fast;

#if defined(PCM) || defined(LOCAL)
    #include "lib/dump.hpp"
#else
    #define dump(...) 42
    #define dump_1d(...) 42
    #define dump_2d(...) 42
    #define cerrendl 42
#endif
//}}}

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