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)

/*author : Yashvardhan
Tuesday, April 07, 2020
8:18 PM
*/

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

using ll = long long;

#define int ll
#define pb push_back
#define vi vector<int>
#define pi pair<int,int>
#define vpi vector<pi>
#define ff first
#define ss second
#define mp make_pair
#define endl '\n'
#define all(a) a.begin(), a.end()
#define initialise(a, x) memset(a, x, sizeof(a))
#define rev(Y) reverse(all(Y))
#define printArr(name,from,to) for(int x = from; x < to; x++) cout << name[x] << " ";


#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _CRT_SECURE_NO_WARNINGS

#ifdef __APPLE__
#define debug(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){cerr << name << " : " << arg1 << endl;}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');
	cerr.write(names, comma - names) << " : " << arg1 << "  ";
	__f(comma + 1, args...);}
#else
#define debug(...)
#endif

template <typename T> ostream& operator << (ostream& os,const vector<T>& v){for(int i=0;i<v.size();++i) os<<v[i]<<" ";return os;} 
template <typename T> ostream& operator << (ostream& os,const set<T>& v){for(auto it:v) os<<it<<" ";return os;} 
template <typename T, typename S> ostream& operator << (ostream& os, const pair<T,S>& v) {os<<v.ff<<" "<<v.ss;return os;}

const int mod = 1e9 + 7;
const int inf = 2e18;
const int ninf = -2e18;

int takemod(int a){return ((a%mod)+mod)%mod;}

int pow(int a, int b, int m)
{
	int ans = 1;
	a %= m;
	while (b)
	{
		if (b & 1)
			ans = (ans * a) % m;
		b /= 2;
		a = (a * a) % m;
	}
	return ans;
}

int modinv(int a){return takemod( pow( takemod(a), mod-2,mod ) );}


signed main()
{

	ios_base::sync_with_stdio(false);
	cin.tie(0);

	#ifdef __APPLE__
		freopen("input.txt","r",stdin);
		freopen("output.txt","w",stdout);
		freopen("error.txt","w",stderr);
	#endif

	time_t t1, t2;
	t1 = clock();
	
	int n,m;
	cin >> n >> m;

	int ans = ((n)*(n-1))/2;
	ans += (m*(m-1))/2;

	cout << ans << endl;
	

	t2 = clock();
	cerr << endl << t2-t1 << endl;
	
	return 0;
}