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>
using namespace std;
 
#define FOR(i, x, n) for(ll i = x; i < n; i++) 
#define pb push_back
#define pf push_front
#define ll long long
#define hii cout << "hii" << endl
#define pii pair<int, int>
#define pll pair<ll, ll>
#define int ll
#define mpp make_pair
#define endl '\n'
#define ff first 
#define ss second
#define vi vector<int>
#define all(s) s.begin(), s.end()
#define si size()

const int N = 1e5 + 5;
double arr[N];

int solve()
{
	int n;
	cin >> n;
	n--;
	cout << (n * (n + 1)) / 2;
}

int32_t main()
{
	ios_base:: sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	// #ifndef ONLINE_JUDGE
 //    freopen("input.txt", "r", stdin);
 //    freopen("output.txt", "w", stdout);
	// #endif
	solve();
	return 0;
}