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 ALL(x) (x).begin(),(x).end()
#define COUT(x) cout<<(x)<<"\n"
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define REP(i, n) for(int i=0;i<n;i++)
#define YES(x) cout<<(x?"YES":"NO")<<"\n"
#define Yes(x) cout<<(x?"Yes":"No")<<"\n"
#define dump(x) cout<<#x<<" = "<<(x)<<"\n"
#define endl "\n"
using G = vector<vector<int>>;
using M = map<int,int>;
using P = pair<int,int>;
using PQ = priority_queue<int>;
using PQG = priority_queue<int,vector<int>,greater<int>>;
using V = vector<int>;
using ll = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

int main() {
  IOS;
  int n; cin >> n;
  COUT( (1+n-1)*(n-1)/2 );
  return 0;
}