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>
using namespace std;
typedef long long ll;
bool isPerfectsquare(ll n)
{
    ll x=sqrt(n);
    return x*x==n;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);

    ll n,ans;
    cin>>n;

    for(ll i=1;i<=10000000;i++)
    {
        if(isPerfectsquare(1ll+8*n*i))
        {
            ll r=sqrt(1ll+8*n*i);
            if((r-1)%2==0){
            ans=(r-1)/2;break;
            }
        }
    }
    cout<<ans<<"\n";
}