Problem:
Score : 400 points

Problem Statement

For an integer N, we will choose a permutation \{P_1, P_2, ..., P_N\} of \{1, 2, ..., N\}.

Then, for each i=1,2,...,N, let M_i be the remainder when i is divided by P_i.

Find the maximum possible value of M_1 + M_2 + \cdots + M_N.

Constraints

N is an integer satisfying 1 \leq N \leq 10^9.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible value of M_1 + M_2 + \cdots + M_N.

Sample Input 1

2

Sample Output 1

1

When the permutation \{P_1, P_2\} = \{2, 1\} is chosen, M_1 + M_2 = 1 + 0 = 1.

Sample Input 2

13

Sample Output 2

78

Sample Input 3

1

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (ll i = 0; i < n; ++i)
#define sz(x) int(x.size())
#define ALL(c) (c).begin(), (c).end()
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())
#define FORV(i, v) for (auto i = v.begin(); i != v.end(); i++)
using namespace std;

using ll = int64_t;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
const int INF = 1'001'001'001;
const ll INFL = (1LL<<60);
const double eps = (1e-9);

const ll mod = 1e9 + 7;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;

  ll ans = 0;
  for (int i = 1; i < n; i++) {
    ans += i;
  }

  cout << ans << endl;

  return 0;
}