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>
#define LL long long
#define REP(i,n) for(int i=0;i<(n);++i)
#define PER(i,n) for(int i=n-1;i>=0;--i)
#define REPA(i,n) for(int i=1;i<(n);++i)
#define foreach(i, n) for(auto &i:(n))
#define PII pair<int,int>
#define PLI pair<long long, int>
#define PLL pair<long long, long long>
#define MOD ((LL)1e9+7) 
#define INF ((LL)1e9+7)
#define INFLL ((LL)1e18)
#define ALL(x) (x).begin(),(x).end()
#define BIT(x) (1LL << (x))
using namespace std;
template<class t, class u> bool chmax(t &a, u b){if(a<b){a=b;return true;}return false;}
template<class t, class u> bool chmin(t &a, u b){if(a>b){a=b;return true;}return false;}

LL modpow(LL x, LL b){
	LL res = 1;
	while(b){
		if(b&1)res = res * x % MOD;
		x = x * x % MOD;
		b>>=1;
	}
	return res;
}

LL modinv(LL x){
	return modpow(x, MOD-2);
}

int main(){
	LL N;
	cin >> N;
	cout << (N-1) * N / 2 << endl;
	return 0;
}