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)

    /* HARDWORK IS THE KEY TO SUCCESS.
       RATING IS TEMPORARY , KNOWLEDGE IS PERMANENT.!!!*/
    #include <bits/stdc++.h>
    #define pragma
    using namespace std;
    #define ll long long
    #define ld long double
    #define T ll t = 1; cin>>t; while(t--)
    #define vin ll n; cin>>n; vector<ll>x(n); for(ll i=0; i<n; i++) cin>>x[i]
    #define vout for(ll i=0; i<n; i++) cout<<x[i]<<" "
    #define ain ll n,k,i; cin>>n; ll a[n]; for( i=0; i<n; i++) cin>>a[i]
    #define aout for(ll i=0; i<n; i++) cout<<a[i]<<" "
    #define matin ll m,n; cin >> n >> m; ll a[n + 1][m + 1] ; for(ll i=1;i<=n;i++) for(ll j=1;j<=m;j++) cin>>a[i][j];
    #define matout for(ll i=0;i<m;i++){for(ll j=0;j<n;j++) cout<<a[i][j]<<" "; cout<<"\n";}
    #define sin set<ll>s; ll n; cin>>n; ll tmp; for(ll i=0;i<n;i++){cin>>tmp; s.I(tmp);}
    #define sout for(auto it=s.begin();it!=s.end();++it) cout<<*it<<" "
    #define F first
    #define S second
    #define I insert
    #define pb push_back
    #define ppb pop_back
    #define AISORT sort(a,a+n)
    #define ADSORT sort(a,a+n,greater<ll>())
    #define VISORT sort(x.begin(),x.end())
    #define VDSORT sort(x.begin(),x.end(),greater<ll>())

    const ll maxs = 1e5+3;
    const ll maxs1 = 1e3+4;
    const ll mod = 1e9 + 7;

    //MODULO OPEARTIONS
    inline ll add(ll a, ll b) { return (mod + a + b) % mod; }
    inline ll sub(ll a, ll b) { return (a - b + mod) % mod; }
    inline ll mul(ll a, ll b) { return (1ll * a * b) % mod; }

    //POWER
    ll fastpow(ll x,ll y)
    {ll res = 1;x = x % mod;while (y > 0){ if (y & 1)res = (res * x) % mod; y = y >> 1 ; x = (x * x) % mod; } return res;}

    //MODULAR INVERSE
    ll inv(ll x){ return fastpow(x , mod - 2);}

//     ll n;
//     ll a[maxs];
//     ll tree[maxs];
    bool vis[maxs];
    vector<ll>graph[maxs];
    //ALWAYS CLEAR THEM WHEN TESTCASES ARE GIVEN
    //ALWAYS RETURN WHEN ST == ED CONDITION IN SEGMENT TREES


    int main()
    {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
      ll n,m;
        cin >> n >> m;
        if(n < 2 && m < 2)
        {
            cout << 0 ;
            return 0;
        }
            
        ll ans = n * (n - 1) / 2;
        ll ANS = m * (m - 1) / 2;
        cout << ans + ANS;
    }