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>
using namespace std;
 
#define cinsc ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(0);
#define all(cont) cont.begin(), cont.end()
#define rall(cont) cont.end(), cont.begin()
#define FOR(i,k,n)  for(int64 i=k;i<(int64)n;i++)
#define FORr(i,k,n)  for(int64 i=(int64)k;i>=n;i--)
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define FOREACHr(it, l) for (auto it = l.rbegin(); it != l.rend(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define PB push_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define read(type) readInt<type>()
#define SQ(x) ((x)*(x))
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int int32;
typedef unsigned long int uint32;
typedef long long int int64;
typedef unsigned long long int  uint64;
 
//ONLINE_JUDGE   LOCAL
 
int64 powr(int64 x,int64 y,int64 p)  
{  
    int res = 1;    
    x = x % p; 
    if (x == 0) return 0;
    while (y > 0){   
        if (y & 1)  
            res = (res*x) % p;   
        y = y>>1; 
        x = (x*x) % p;  
    }  
    return res;  
} 
 
void  in_o(){
#ifndef LOCAL
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
#endif
	cinsc;
}


int main(){
//	in_o();
	int64 n;
	cin>>n;
	cout<<n*(n-1)/2;
	return 0;
}