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 with AC Library v1.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <random>
#include <chrono>
#include <algorithm>
#define rep(i, x) for(ll i = 0; i < x; i++)
#define CONCAT_(x,y) x##y
#define CONCAT(x,y) CONCAT_(x,y)
#define _v1 CONCAT(qyw1, __LINE__) // unique name in same line (otherwise use __COUNTER__)
#define _v2 CONCAT(qyw2, __LINE__)
#define forbits(i, m) for (int _v1 = m, i = __builtin_ctz(_v1); _v1 != 0; _v1 ^= _v1 & -_v1, i = __builtin_ctz(_v1))
#define forsubm(s, m) for (int s = m; s >= 0; s = !s ? -1 : m & (s - 1))
int __nextXZW(int v){int t=v|(v-1);int w=(t+1)|(((~t&-~t)-1)>>(__builtin_ctz(v)+1));return w;}
#define ksets(i, n, k) for (int i = (1u << k) - 1, max = i << (n - k); i <= max; i = __nextXZW(i))
using namespace std;
typedef long long ll;
const ll MOD = (ll)1e18+7;
const ll N = 5e5+5, inf = 1e9+5000;
ll add(ll x, ll y) { x += y; if (x >= MOD) return x - MOD; return x; }
ll sub(ll x, ll y) { x -= y; if (x < 0) return x + MOD; return x; }
ll mult(ll x, ll y) { return (x * y) % MOD; }
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
#include <atcoder/all>
using namespace atcoder;

ll get_k(ll x, ll y) {
  auto res = crt({0, y-1}, {x, y});
  return (res.first != 0 && 0 != res.second) ? res.first : inf;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  
  ll n; cin >> n;
  n *= 2;
  ll best_k = inf;
  for (ll x = 1; x*x <= n; x++) {
    if (n % x == 0) {
      best_k = min(best_k, get_k(x, n/x));
      if (n/x != x) {
        best_k = min(best_k, get_k(n/x, x));
      }
    }
  }

  cout << best_k << endl;

  return 0;
}