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 REPP(i, n) for(int i = 1; i <= n; ++i)
#define REPR(i, n) for(int i = n-1; i >= 0; --i)
#define FOR(i, m, n) for(int i = m; i < n; ++i)
#define ALL(v) v.begin(), v.end()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
const ll MOD = 1000000007;
const ll INF = 1000000000;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b) {if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b) {if(a<b) a=b;}

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