Problem:
Score : 600 points

Problem Statement

Given is an integer N.
Find the minimum possible positive integer k such that (1+2+\cdots+k) is a multiple of N.
It can be proved that such a positive integer k always exists.

Constraints

1 \leq N \leq 10^{15}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer in a line.

Sample Input 1

11

Sample Output 1

10

1+2+\cdots+10=55 holds and 55 is indeed a multple of N=11.
There are no positive integers k \leq 9 that satisfy the condition, so the answer is k = 10.

Sample Input 2

20200920

Sample Output 2

1100144

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h> 
#include <chrono>
using namespace std;
 
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
 
 
//typedef
//------------------------------------------
typedef pair<int, int> PII;
typedef long long LL;
typedef pair<LL, LL> PLL;
 
//container util
//------------------------------------------
#define PB emplace_back
#define MP make_pair 
#define SZ(a) int((a).size())
//repetition
//------------------------------------------
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define SORT(c) sort((c).begin(),(c).end())
#define ALL(a)  (a).begin(),(a).end()
//constant
//--------------------------------------------
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))


int main(){
    LL N;
    cin>>N;
    for(LL i=1;i*i<=2*N;i++){
        if(i*(i+1LL)%(2*N)==0){
            cout<<i<<endl;return 0;
        }
    }
    cout<<N-1<<endl;
    return 0;
}