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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <unistd.h>
#include <sys/time.h>
#include <stdlib.h>
#include <atcoder/math>

using namespace std;
using namespace atcoder;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<P, ll> T;
typedef pair<long double , ll> Ps;
typedef pair<ll, bool> Pb;
typedef pair<ll, vector<ll>> Pd;

const ll INF = 3e18;
const ll fact_table = 3200008;
long double Pi = 3.1415926535897932384626;

priority_queue <ll> pql;
priority_queue <P> pqp;
priority_queue <P> bag;
//big priority queue
priority_queue <ll, vector<ll>, greater<ll> > pqls;
priority_queue <P, vector<P>, greater<P> > pqps;
//small priority queue
//top pop

ll dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
ll dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
//↓,→,↑,←


#define endl "\n"
#ifdef ENJAPMA
#undef endl
#endif

#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)();
#define pc(x) cout << x << ",";
#define rep(i, n) for (ll i = 0; i < (n); i ++)

typedef vector<ll> vec;
typedef vector<vector<ll>> mat;

const ll mod = 998244353ll;
// const ll mod = 1000000007ll;

ll mypow(ll a, ll b, ll m = mod) {ll x = 1; while (b) {while (!(b & 1)) {(a *= a) %= m; b >>= 1;}(x *= a) %= m; b--;} return x;}
vec readvec(ll read) { vec res(read); for (int i = 0; i < read; i++) { cin >> res[i]; } return res;}
void YES(bool cond) { if (cond) { p("YES");} else { p("NO");} return;}
void Yes(bool cond) { if (cond) { p("Yes");} else { p("No");} return;}
void line() { p("--------------------"); return;}

/*
ll fact[fact_table + 5], rfact[fact_table + 5];

void c3_init() {
    fact[0] = rfact[0] = 1;
    for (ll i = 1; i <= fact_table; i++) {
        fact[i] = (fact[i - 1] * i) % mod;
    }
    rfact[fact_table] = mypow(fact[fact_table], mod - 2, mod);
    for (ll i = fact_table; i >= 1; i--) {
        rfact[i - 1] = rfact[i] * i;
        rfact[i - 1] %= mod;
    }
    return;
}
ll c3(ll n, ll r) {
    return (((fact[n] * rfact[r]) % mod ) * rfact[n - r]) % mod;
}
*/

struct Timer {
    int64_t start; const int64_t CYCLES_PER_SEC = 2800000000;
    Timer() { reset(); }
    void reset() { start = getCycle(); }
    inline double get() { return (double)(getCycle() - start) / CYCLES_PER_SEC; }
    inline int64_t getCycle() {
        uint32_t low, high; __asm__ volatile ("rdtsc" : "=a" (low), "=d" (high));
        return ((int64_t)low) | ((int64_t)high << 32);
    }
};

bool multicase = false;

ll mul(ll a, ll b) {
    if (b == 0) return 0;
    if (a > INF / b) {
        return INF;
    } else {
        return a * b;
    }
}

void solve() {
    ll n;
    cin >> n;
    if (n == 2) {
        p(3);
        return;
    }
    vec v;
    for(int i=2;i*i<=n;i++){
        ll res = 1ll;
        if (n % i == 0) {
            while(n % i == 0) {
                res *= i;
                n /= i;
            }
        }
        if (i == 2) res *= 2;
        if (res != 1ll) v.pb(res);
    }
    if (n != 1) v.pb(n);
    ll m = v.size();
    ll mbit = (1ll << m);
    ll ans = INF;
    for(int i=0;i<mbit;i++){
        ll ka = 1ll, kb = 1ll;
        for(int j=0;j<m;j++){
            if (i & (1ll << j)) {
                ka *= v[j];
            } else {
                kb *= v[j];
            }
        }
        ll x = inv_mod(ka, kb);
        x = kb - x;
        ll res = mul(x, ka);
        ans = min(ans, res);
    }
    p(ans);
    return;
}

int main() {
    // init();
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    ll q, testcase = 1;
    if (multicase) {
        cin >> q;
    } else {
        q = 1;
    }
    while (q--) {
        // pu("Case ");pu("#");pu(testcase);pu(": ");
        solve();
        testcase++;
    }

    // solve();
    return 0;
}