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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
#define rep(i, n) for(int i = 0; i < n; i++)
#define REP(i, a, b) for(int i = a; i <= b; i++)
#define all(x) x.begin(), x.end()
#define fi first
#define se second
#define debug(...) cerr<<__LINE__<<" ["<<#__VA_ARGS__<<"]:",debug_out(__VA_ARGS__)
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec)
{ for (auto v : vec) os << v << ' '; os << '\n'; return os; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa)
{ os << '(' << pa.first << ',' << pa.second << ')'; return os; }
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail> void debug_out(Head H, Tail... T)
{ cerr << " " << H; debug_out(T...); }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = (1<<30) - 1;
const ll LINF = (1LL<<60) - 1;

void solve() {
    ll n;
    cin >> n;
    cout << n * (n - 1) / 2 << endl;
}

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