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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define pp pair<pair<ll, ll>,pair<ll, ll>>
#define pll pair<ll,ll>
#define pdd pair<double,double>
#define vii vector<int>
#define vll vector<ll>
#define mat vector<vector<ll>>
#define lb lower_bound
#define ub upper_bound
#define pb push_back
#define eb emplace_back
#define fi first
#define sc second
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep2(i,a,b) for(ll i=a;i<b;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define all(x) x.begin(),x.end()
#define sz(x) (ll) (x).size()
#define pq priority_queue<ll>
#define pqg priority_queue<ll,vector<ll>,greater<ll>>
#define LB(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define UB(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
#define ERASE(v) sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end())
#define int ll
// #define ll int
using namespace std;
const ll INF = (1 << 30 ) - 1;
const ll LLINF = (1LL << 60LL);
const ll MOD = 1000000007;
const ll mod = 998244353;
const ll MAX = 1100000;
const double pi = acos(-1);
const double eps = 1e-10;
ll dx[8] ={1,0,-1,0,1,-1,1,-1};
ll dy[8] ={0,1,0,-1,1,-1,-1,1};
 
template<class T> 
inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> 
inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
 
struct Timer{
    chrono::system_clock::time_point start, end;
    Timer(){ start = chrono::system_clock::now(); }
    ~Timer(){
        end = chrono::system_clock::now();
        auto msec = chrono::duration_cast<chrono::milliseconds>(end - start).count();
        cerr<<"time : "<<msec<<" ms"<<endl;
    }
};


void solve(){
    ll n,m;
    cin>>n>>m;
    cout<<n*(n-1)/2+m*(m-1)/2<<endl;



}


signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    // Timer TM;
     
    solve();


    return 0;
}