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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define rep(i,j,k) for(int i=j;i<k;i++)
#define all(x) (x).begin(),(x).end()
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define pb push_back
#define mp make_pair
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
const int  COMMAX = 510000;

long long fac[COMMAX], finv[COMMAX], inv[COMMAX];

int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcd(int a,int b){return a*b / gcd(a,b);}
// テーブルを作る前処理
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < COMMAX; i++){
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

// 二項係数計算
long long COM(int n, int k){
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

struct UnionFind {
  vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
  vector<int> rank;
  int num;
  UnionFind(int N) : par(N) , rank(N) { //最初は全てが根であるとして初期化
    num = N;
    for(int i = 0; i < N; i++){
      par[i] = i;
      rank[i] = 0;
    }
  }

  int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
    if (par[x] == x) return x;
    return par[x] = root(par[x]);
  }

  void unite(int x, int y) { // xとyの木を併合
    int rx = root(x); //xの根をrx
    int ry = root(y); //yの根をry
    if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
    if(rank[rx] < rank[ry]){
      par[rx] = ry;  
    }else{
      par[ry] = rx;
      if(rank[rx] == rank[ry])rank[rx]++;
    }
    num--;
  }

  bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
    return root(x) == root(y);
  }
};

const int MS = 2;
const long long mod[] = {999999937LL, 1000000007LL}, base = 9973;
struct rolling_hash {
    int n;
    vector<long long> hs[MS], pw[MS];
    rolling_hash(){}
    rolling_hash(const string &s) {
        n = s.size();
        for (int i = 0; i < MS; i++) {
            hs[i].assign(n+1,0);
            pw[i].assign(n+1,0);
            hs[i][0] = 0;
            pw[i][0] = 1;
            for (int j = 0; j < n; j++) {
                pw[i][j+1] = pw[i][j]*base%mod[i];
                hs[i][j+1] = (hs[i][j]*base+s[j])%mod[i];
            }
        }
    }

    long long hash(int l, int r, int i) { return ((hs[i][r]-hs[i][l]*pw[i][r-l])%mod[i]+mod[i])%mod[i]; }

    bool match(int l1, int r1, int l2, int r2) {
        bool ret = 1;
        for (int i = 0; i < MS; i++) ret &= hash(l1,r1,i)==hash(l2,r2,i);
        return ret;
    }

    bool match(int l, int r, long long h[]) {
        bool ret = 1;
        for (int i = 0; i < MS; i++) ret &= hash(l,r,i)==h[i];
        return ret;
    }
};

int main(){
  int n,m;
  int res = 0;
  cin >> n >> m;
  while(n>0){
    n--;
    res += n;
  }
  while(m>0){
    m--;
    res += m;
  }
  cout << m << endl;
  return 0;
}