Problem: Now you get Baby Ehab's first words: "Given an integer n, find the longest subsequence of [1,2, …, n-1] whose product is 1 modulo n." Please solve the problem.

A sequence b is a subsequence of an array a if b can be obtained from a by deleting some (possibly all) elements. The product of an empty subsequence is equal to 1.

Input

The only line contains the integer n (2 ≤ n ≤ 10^5).

Output

The first line should contain a single integer, the length of the longest subsequence.

The second line should contain the elements of the subsequence, in increasing order.

If there are multiple solutions, you can print any.

Examples

Input


5


Output


3
1 2 3 

Input


8


Output


4
1 3 5 7 

Note

In the first example, the product of the elements is 6 which is congruent to 1 modulo 5. The only longer subsequence is [1,2,3,4]. Its product is 24 which is congruent to 4 modulo 5. Hence, the answer is [1,2,3].

Solution (in CPP):

Solution:
// DeNsE - EcLiPsE //
// WHAT is DEAD may NEVER die //

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
#include <map>
#include <unordered_map>
#include <cmath>
#include <iomanip>
#include <set>
#include <cstring>
#include <stack>
#include <sstream>
#include <queue>
#include <unordered_set>
#include <cstdlib>

using namespace std;

#define nitro ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define int long long
#define double long double
//#define endl " \n"

const int inf = (1ll << 62) - 1;
const int mod = 1e9 + 7;
const int N = 2e5 + 2;
const double pi = 3.14159265358979323846;

int bPow(int a, int b){
    int res = 1;
    while(b) {
        if (b & 1) {
            res = (res * a) % mod;
        }
        b >>= 1;
        a = (a * a) % mod;
    }
    return res % mod;
}

int gcd(int a, int b)
{
    if(a < b)
        swap(a, b);
    if (b == 0)
        return a;
    return gcd(b, a % b);

}

vector<int> fact(N, 0ll);
void factorial(){
    fact[0] = 1, fact[1] = 1;
    for(int i = 2; i < N; i++){
        fact[i] = (fact[i - 1] * i) % mod;
    }
}

int ncr(int n, int r){
    if(r > n)
        return 0;
    int ans = fact[n] % mod;
    ans *= bPow(fact[r], mod - 2) % mod;
    ans %= mod;
    ans *= bPow(fact[n - r], mod - 2) % mod;
    ans %= mod;
    return ans;
}

vector<int> primes(N, -1);;
void sieve(){
    iota(primes.begin(), primes.end(), 0);
    for (int i = 2; i * i <= N; ++i) {
        if(primes[i] == i){
            for (int j = 2 * i; j < N; j += i) {
                primes[j] = i;
            }
        }
    }
}


void solve() {
    int n;
    cin >> n;
    vector<bool> cnt(n + 1, false);
    int product = 1;
    for (int i = 1; i <= n; ++i) {
        if(gcd(n, i) == 1){
            cnt[i] = true;
            product *= i;
            product %= n;
        }
    }
    if(product != 1)
        cnt[product] = false;
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
        if(cnt[i])
            ans.push_back(i);
    }
    cout << ans.size() << endl;
    for(int an : ans)
        cout << an << " ";
    cout << endl;
}

signed main(){
    nitro
    sieve();
    int tc = 1, test = 1;
    //cin >> tc;
    while(tc--){
        //cout << "Case #" << test++ << ": ";
        solve();
    }
}