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>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define rep1(i,n) for(int i=1; i<=(n); ++i)
#define repi(i,a,b) for(int i=a; i<=(b); ++i)
#define rrep(i,n) for(int i=(n-1); i>=0; --i)
#define ALL(obj) (obj).begin(), (obj).end()
#define RALL(obj) (obj).rbegin(), (obj).rend()
#define pb push_back
#define mp make_pair
#define to_s to_string
#define sz(v) (int)v.size()
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() )
#define pr(x) cout<<(x)<<'\n'
#define debug(x) cout << #x << ": " << (x) << '\n'
#define yes "Yes"
#define no "No"
using namespace std;
using ll = long long;
using Edge = pair<int,long long>;
using Graph = vector<vector<Edge>>;
typedef pair<int,int> P;
struct aaa{aaa(){ cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(20);};}aaaaaaa;
int MOD = 1e9 + 7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return (a*b)/gcd(a,b);}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

int main(){
    int N; cin >> N;
    pr(N*(N-1)/2);
}