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 <iostream>
#include <array>
#include <algorithm>
#include <vector>
#include <set>
#include <unordered_set>
#include <cmath>
#include <complex>
#include <deque>
#include <iterator>
#include <numeric>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <string>
#include <tuple>
#include <utility>
#include <limits>
#include <iomanip>
using namespace std;

using ll=long long;
template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
#define rep(i, k, n) for (ll i=k; i<(ll)n; ++i)
#define REP(i, n) rep(i, 0, n)
#define ALL(v) v.begin(),v.end()
template<class T> inline bool chmax(T& a, T b) {if (a<b) {a=b; return true;} return false;}
template<class T> inline bool chmin(T& a, T b) {if (a>b) {a=b; return true;} return false;}

const ll MOD = 1000000007;
const ll HIGHINF = (ll)1e18;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n; cin >> n;
    cout << n*(n-1)/2 << '\n';
    return 0;
}