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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define el				'\n'
#define sz(v)			((int)((v).size()))
#define all(v)          ((v).begin()),((v).end())
#define clr(v, d)		memset(v, d, sizeof(v))

double const EPS = 1e-8, PI = acos(-1);
const int N = 2e5 + 9, M = 50 + 9, OO = (int) 1e9 + 1;
const long long MOD = 1e9 + 7;

typedef long long ll;

void OUTPUT() {
	cout << fixed << setprecision(12);
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
}

void INPUT() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
#endif
	cout << fixed << setprecision(12);
//	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
}

vector<vector<int>> edges(N);
int n, q, x, y;
ll counter[N];
bool vis[N];

int main() {
	OUTPUT();
//	INPUT();
	cin >> n;
	x = n;
	x--;
	ll ans = ((1 + x) / 2.0) * x;
	cout << ans;
//	for (int i = 1; i <= n; i++) {
//		y += i % x;
//		cout << i % x << " " << i << el;
//		x--;
//	}
//	cout << y;

	return 0;

}