1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef unsigned long long LL;
19#define pi pair < int ,int >
20#define MP make_pair

 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6struct Mat
 7{
 8    LL mat[2][2];
 9    void clear()
10    {
11    ms(mat,0);
12    }
13    void pr()
14    {
15    for ( int i = 0 ; i < 2 ; i++)
16        for ( int j = 0 ; j < 2 ; j++)
17        printf("%lld%c",mat[i][j],j==1?'\n':' ');
18    }
19}M,M1;
20const Mat P = {1,1,1,0};
21Mat mul (Mat a,Mat b,LL mod)
22{
23    Mat c;
24    c.clear();
25    for ( int i = 0 ; i < 2 ; i++)
26    for ( int j = 0 ; j < 2 ; j++)
27        for ( int k  = 0 ; k < 2 ; k++)
28        c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j]%mod)%mod;
29    return c;
30}
31Mat mat_ksm(Mat a,LL b,LL mod)
32{
33    Mat res;
34    res.clear();
35    for ( int i = 0 ; i < 2 ; i++) res.mat[i][i] = 1;
36    while (b>0)
37    {
38    if (b&1) res = mul(res,a,mod);
39    b = b >> 1LL;
40    a = mul(a,a,mod);
41    }
42    return res;
43}
44LL gcd(LL a,LL b)
45{
46    return b?gcd(b,a%b):a;
47}
48const int N = 5E6+7;
49bool prime[N];
50int p[N];
51int pri_tot;
52void Lineisprime() //换成线性晒了。
53{
54   // int cnt = 0 ;
55    ms(prime,true);
56    for ( int i = 2 ; i < N ; i++)
57    {
58    if (prime[i]) p[pri_tot++] = i ;
59    for ( int j =  1 ;  j < pri_tot && i*p[j] < N ; j++)
60    {
61        prime[i*p[j]] = false;
62        if (i%p[j]==0) break;
63    }
64    }
65}

 1LL ksm( LL a,LL b,LL mod)
 2{
 3   LL res = 1;
 4   while (b>0)
 5   {
 6       if (b&1) res = (res * a) % mod;
 7       b = b >> 1LL;
 8       a = a * a % mod;
 9   }
10   return res;
11}
12LL legendre(LL a,LL p) //勒让德符号,判断二次剩余
13{
14    if (ksm(a,(p-1)>>1,p)==1) return 1;
15    return -1;
16}

 1LL pri[N],num[N];//分解质因数的底数和指数。
 2int cnt; //质因子的个数
 3void solve(LL n,LL *pri,LL *num)
 4{
 5    cnt = 0 ;
 6    for ( int  i = 0 ; 1LL*p[i] * p[i] <= n  && i < pri_tot ; i++)
 7    {
 8    if (n%p[i]==0)
 9    {
10        int Num = 0 ;
11        pri[cnt] = p[i];
12        while (n%p[i]==0)
13        {
14        Num++;
15        n/=p[i];
16        }
17        num[cnt] = Num;
18        cnt++;
19    }
20    }
21    if (n>1)
22    {
23    pri[cnt] = n;
24    num[cnt] = 1;
25    cnt++;
26    }
27}
28LL fac[N];
29int cnt2; //n的因子的个数
30void get_fac(LL n)//得到n的所有因子
31{
32    cnt2 = 0 ;
33    for (int i =  1 ; i*i <= n ; i++)
34    {
35    if (n%i==0)
36    {
37        if (i*i!=n)
38        {
39        fac[cnt2++] = i ;
40        fac[cnt2++] = n/i;
41        }
42        else fac[cnt2++] = i;
43    }
44    }
45}
46LL delta;
47const LL LOOP[10]={3,8,20};
48LL ask_loop(int id) //我好傻啊。。并不一定所有因子都有啊。。。
49{
50    return LOOP[id];
51}
52LL find_loop(LL n)
53{
54    //cout<<"n:"<<n<<endl;
55    solve(n,pri,num);
56    //puts("pri:");
57   // for ( int i = 0 ; i < cnt ; i++)  printf("i:%d %lld\n",i,pri[i]);
58    LL ans = 1;
59    //cout<<"CNT:"<<cnt<<endl;
60    for ( int i = 0 ; i < cnt ; i++)
61    {
62    LL rec = 1;
63    if (pri[i]==2) rec =  3;
64    else if (pri[i]==3) rec = 8;
65    else if (pri[i]==5) rec = 20;
66    else
67    {
68        if (legendre(5,pri[i])==1)
69        get_fac(pri[i]-1);
70        else get_fac((pri[i]+1LL)*(3-1)); //为什么可以假设循环节不大于2*(p+1)???
71        sort(fac,fac+cnt2);
72       // for  ( int qqq = 0; qqq < cnt2 ; qqq++) printf("fac: %lld  ",fac[qqq]);
73        for ( int j = 0 ; j < cnt2 ; j++) //挨个验证因子
74        {
75        Mat tmp = mat_ksm(M,fac[j],pri[i]); //下标从0开始，验证fac[j]为循环节，应该看fib[0]==fib[fac[j]]和fib[1]==fib[fac[j]+1]是否成立
76        tmp = mul(tmp,M1,pri[i]);
77        if (tmp.mat[0][0]==1&&tmp.mat[1][0]==0)
78        {
79            rec = fac[j];
80            break;
81        }
82        }

 1    }
 2    for ( LL j = 1 ; j < num[i] ; j++)
 3        rec *=pri[i];
 4    ans = ans / gcd(ans,rec) * rec;
 5    }
 6    return ans;
 7}
 8void init()
 9{
10    M.clear();
11    M.mat[0][0] = M.mat[0][1] = M.mat[1][0] = 1;
12    M1.clear();
13    M1.mat[0][0] = 1;
14}
15LL n;
16int main()
17{

1    init();
2    Lineisprime();
3    while (~scanf("%llu",&n))
4    {
5      //  cout<<"n:"<<n<<endl
6        if (n==2) puts("3");
7        else 
8        printf("%llu\n",find_loop(n)/2);

1#ifndef ONLINE_JUDGE  
2    fclose(stdin);
3#endif
4    return 0;
5}