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>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define FOR(i,j,n) for(int i=(j);i<=((int)n);++i)
#define N 1000005
#define MOD 1000000007
#define all(c) (c).begin(),(c).end()
#define tr(container, it) for(__typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define present(container, element) (container.find(element) != container.end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define pb push_back
typedef long long int ll;
typedef std::vector< int > vi;
typedef std::vector< vi > vvi;
typedef std::pair< int,int > ii;
using namespace std;

vector <int>g[N];
void init()
{
	for(int i = 0; i < N; i++) {
		g[i].clear();
	}
}


int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n;
    cin>>n;
    cout<<(n*(n-1))/2<<endl;
   
}