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)

/*g++ main.cpp -o main.out*/
/*./main.out*/

#include<bits/stdc++.h>
using namespace std;

#define LL long long
#define LD long double
#define Mod 1000000007
#define L_Mod 17100000013

const double eps = 1e-9;
const int INF = 0x3f3f3f3f;
const double PI = 3.1415926535;

#define NO_TLE std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define ree(a,n) memset(a,n,sizeof(a));

#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define red(i,l,u) for(int (i)=(int)(l);(i)>=(int)(u);--(i))

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