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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <tuple>
#include <cstdint>
#include <cstdio>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <cctype>

using namespace std;

#define ll long long
#define ld long double

#define rep(i, n)      for (int i = 0; i < (int)(n); ++i)
#define rep3(i, m, n)  for (int i = (m); i < (int)(n); ++i)
#define rrep(i, n)     for (int i = (int)(n)-1; i >= 0; --i)
#define rrep3(i, m, n) for (int i = (int)(n)-1; i >= (m); --i)
#define all(x) begin(x), end(x)
#define rall(x) end(x), begin(x)

#define endl '\n'


ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

ll solve(ll N) {
  return (int)((double)N*(N-1)/2+0.5);
}

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

  ll N;
  cin >> N;

  auto ans = solve(N);
  cout << ans << endl;

  return 0;
}