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)

#include<bits/stdc++.h>
#include<cstring>
using namespace std;
//#define num 1000000007
#define num 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define up(a,b) (((a)%(b)==0)?((a)/(b)):((a)/(b))+1)
#define inarr(i,a,n,A) for(ll i=a;i<n;i++) { cin>>A[i]; }
#define vll vector<long long>
#define pii pair <int,int>
#define pll pair<long long, long long>
#define vpii vector< pair <int,int>>
#define vpll vector< pair<long long, long long>>
#define max(a,b) ((a>b)?a:b)
#define min(a,b) ((a>b)?b:a)
#define max3(a,b,c) (a>b)?(a>c)?a:c:(b>c)?b:c
#define min3(a,b,c) (a<b)?(a<c)?a:c:(b<c)?b:c
#define avg(a,b) ((a)+(b))/2
#define push push_back
#define pop pop_back
#define mp make_pair
#define sf int flag=0     //Remember semicolon
#define assmin(a,b) if(b<=a) { a=b; }
#define assmax(a,b) if(b>=a) { a=b; }
#define fi first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
typedef long long ll;
 
ll mod(ll a, ll b)
{
	if(a%b<0)
	{
		return a%b + b;
	}
	return a%b;
}
 
 
ll mod_exp(ll a, ll b, ll c)
{
	ll res=1;
	a=a%c;
	while(b>0)
	{
		if(b%2==1)
		{
			res=(res*a)%c;
		}
		b/=2;
		a=(a*a)%c;
	}
	return res;
}
ll gcd(ll a, ll b)
{
    if(b==0)
    {
        return a;
    }
    return gcd(b, a%b);
}
bool prime[3001];
void SieveOfEratosthenes() 
{ 
    memset(prime, true, sizeof(prime)); 
    prime[1]=false;
    for (int p=2; p*p<=3000; p++) 
    { 
        if (prime[p] == true) 
        { 
            for (int i=p*p; i<=3000; i += p) 
                prime[i] = false; 
        } 
    } 
}
ll powe[20];
void power()
{
	powe[0]=1;
	REP(i,1,20)
	{
		powe[i]=(powe[i-1]*10)%num;
        if(powe[i]<0)
        powe[i]+=num;
	}
}
int main()
{
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	ll n,m;
	cin>>n>>m;
	cout<< (n*(n-1) + m*(m-1))/2;
	return 0;
}