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>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<cstring>
#include<bitset>
#include<stack>

#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define ALL(x) x.begin(),x.end()
#define mst(var,val) memset(var,val,sizeof(var))
#define IOS ios::sync_with_stdio(false);cin.tie(0)
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
#define ls rt<<1
#define rs rt<<1|1
#ifdef local
#define dbg(a) cout << #a": " << a << endl;
#else
#define dbg(a)
#endif // loacl


using namespace std;
typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <double, double> pdd;
const ll inf=0x3f3f3f3f;
const long long INF=0x3f3f3f3f3f3f3f3fLL;
const double PI=acos(-1.0);
const long double eps=1e-6;
const int mod=1e9+7;
const int maxn=20;
const int N=1e5+10;
const int M=N<<1;
const ll mm=(1LL<<32);

template <class T>
inline void read(T &x) {
    x = 0;
    char c = getchar();
    bool f = 0;
    for (; !isdigit(c); c = getchar()) f ^= c == '-';
    for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    x = f ? -x : x;
}

template <class T>
inline void write(T x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    T y = 1;
    int len = 1;
    for (; y <= x / 10; y *= 10) ++len;
    for (; len; --len, x %= y, y /= 10) putchar(x / y + 48);
}
ll qpow(ll a,ll b,ll MOD)
{
    ll ans=1;
    while(b)
    {
        if(b&1) ans=(ans*a)%MOD;
        b>>=1;
        a=(a*a)%MOD;
    }
    return ans;
}

int a[N];
int main()
{
#ifdef local
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
#endif // local
    ll n;cin>>n;
    cout<<n*(n-1)/2;
}