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 <ext/pb_ds/detail/standard_policies.hpp>
 
const double pi=acos(-1.0);
//memset ( a , 0 , n * sizeof(ll) ) ;
using namespace std;
//using namespace __gnu_pbds; 
#define   endl    '\n'
#define   sl(n)     scanf("%lld",&n)
#define   mp      make_pair
#define   pb      push_back
#define   ppb     pop_back
#define   fi      first
#define   se      second
#define   ll      long long
#define   ull     unsigned long long
#define   ld      long double
#define   pii     pair<int, int>
#define   f(i,a,b)  for(ll i = (ll)(a); i < (ll)(b); i++)
#define   rf(i,a,b)   for(ll i = (ll)(a); i > (ll)(b); i--)
#define   ms(a,b)   memset((a),(b),sizeof(a))
#define   abs(x)    ((x<0)?(-(x)):(x))
#define   MAX     1000005
#define   inf     LLONG_MAX
#define   ninf    LLONG_MIN
#define   MIN     INT_MIN
#define   yeet    return 0;
#define fast_io ios_base::sync_with_stdio (false) ; cin.tie(0) ; cout.tie(0) ;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// Use cout.flush() for interactive problems.
// Use this for increased stack size: g++ -o a.exe -Wl,--stack=256000000 -O2 source.cpp
inline long long  MAX2(long long  a, long long  b){return (a)>(b)?(a):(b);}
inline long long  MAX3(long long  a, long long  b,long long  c){return (a)>(b)?((a)>(c)?(a):(c)):((b)>(c)?(b):(c));}
inline long long  MIN2(long long  a, long long  b){return (a)<(b)?(a):(b);}
inline long long  MIN3(long long  a, long long b,long long c){return (a)<(b)?((a)<(c)?(a):(c)):((b)<(c)?(b):(c));}

//typedef
typedef long int int32;
typedef unsigned long int uint32;
typedef long long int int64;
typedef unsigned long long int  uint64;
 
 
int mod = 1e9 +7 ;
int64_t extGcd(int64_t a, int64_t b, int64_t& x, int64_t& y) {if (!a) {x = 0;y = 1;return b;}int64_t x1, y1;int64_t d = extGcd(b % a, a, x1, y1);x = y1 - (b / a) * x1;y = x1;return d;}

inline int addMod(int a, int b, int m = mod) {return ((int64_t)a + b) % m;}
inline int mulMod(int a, int b, int m = mod) {  return ((int64_t)a * b) % m;}
inline int divMod(int a, int b, int m = mod) {int64_t x, y;int64_t g = extGcd(b, m, x, y);if (g != 1) {cerr << "Bad gcd!" << endl;for (;;);}x = (x % m + m) % m;return mulMod(a, x, m);}
int dx[]={1,1,0,-1,-1,-1, 0, 1};
int dy[]={0,1,1, 1, 0,-1,-1,-1}; 
inline ll exp(ll a,ll b){if(a==0)return 0ll;ll r=1LL;while(b>0){if(b&1){r=r*(a%mod);r=(r+mod)%mod;}b/=2;a=(a%mod)*(a%mod);a=(a+mod)%mod;}return (r+mod)%mod;}
ll gcd(ll a,ll b){if(b==0)return a;if(a==0)return b;return gcd(b,a%b);}
uint32 setbits(ll n){uint32 count=0;while (n){n&=(n-1);count++;}return count; } 
//int clearRangeItoJ(int n,int i,int j){ int ones=(~0); int a= ones<<(j+1);int b=(1<<i)-1;int mask=a|b; int ans=n&mask;return ans}

// ll f[MAX];
// ll iv[MAX];
// ll C(ll n,ll r){
// 	 if(n<r)return 0;
// 	 ll ans=(f[n]%mod * iv[r]%mod)%mod;
// 	 ans=(ans%mod * iv[n-r]%mod)%mod;
// 	 return ans%mod;
// }
// void init(){
// 	 f[0]=1;
//      iv[0]=1;
//      f(i,1,MAX)f[i]=(i*f[i-1])%mod;
//      iv[MAX-1]=exp(f[MAX-1],mod-2);

//      for(int i=MAX-2;i>=0;--i)iv[i]=(iv[i+1]*(i+1))%mod;
	 
// }

////****************************************************************************************************************************************************************************************************************////

 int main(){

 	 ll n;
 	 cin>>n;
 	 ll ans= (n*1LL*(n-1))/2;
 	 cout<<ans<<endl;

	  }